The electromotive force of the cell Cd

The electromotive force of the cell \[ \mathrm{Cd} | \mathrm{CdCl}_{2} \cdot 2 \frac{1}{2} \mathrm{H}_{2} \mathrm{O}, \text { sat. solution }|\mathrm{AgCl}| \mathrm{Ag} \] $\operatorname{at} 25^{\circ} \mathrm{C}$ is $0.67533 \mathrm{V}$, and the temperature coefficient is $-6.5 \times$ $10^{-4} \mathrm{V} \mathrm{K}^{-1} .$ Calculate the values of $\Delta_{\mathrm{r}} G, \Delta_{\mathrm{r}} S,$ and $\Delta_{\mathrm{r}} H$ at $25^{\circ} \mathrm{C}$ for the reaction \[ \mathrm{Cd}(\mathrm{s})+2 \mathrm{AgCl}(\mathrm{s})+2 \frac{1}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{l})=2 \mathrm{Ag}(\mathrm{s})+\mathrm{CdCl}_{2} \cdot 2 \frac{1}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \]

The electromotive force of the cell
\[
\mathrm{Cd} | \mathrm{CdCl}_{2} \cdot 2 \frac{1}{2} \mathrm{H}_{2} \mathrm{O}, \text { sat. solution }|\mathrm{AgCl}| \mathrm{Ag}
\]
$\operatorname{at} 25^{\circ} \mathrm{C}$ is $0.67533 \mathrm{V}$, and the temperature coefficient is $-6.5 \times$ $10^{-4} \mathrm{V} \mathrm{K}^{-1} .$ Calculate the values of $\Delta_{\mathrm{r}} G, \Delta_{\mathrm{r}} S,$ and $\Delta_{\mathrm{r}} H$ at $25^{\circ} \mathrm{C}$
for the reaction
\[
\mathrm{Cd}(\mathrm{s})+2 \mathrm{AgCl}(\mathrm{s})+2 \frac{1}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{l})=2 \mathrm{Ag}(\mathrm{s})+\mathrm{CdCl}_{2} \cdot 2 \frac{1}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{s})
\]

Key Concepts

Entropy (?rS) and Enthalpy (?rH) in Thermodynamics

Entropy change (?rS) measures the change in disorder or randomness during a reaction, while enthalpy change (?rH) represents the heat absorbed or released. Their relationship with Gibbs free energy is given by ?rG = ?rH - T?rS, a fundamental equation that connects electrical measurements in a cell to intrinsic thermodynamic properties of the reaction.

Temperature Dependence of Cell Potential

The temperature coefficient of the cell potential describes how the potential difference changes with temperature. This concept is crucial because the derivative dE/dT can be used to relate changes in the cell potential to the reaction’s entropy change, providing insight into the thermal properties of the system.

Gibbs Free Energy Change (?rG)

The Gibbs free energy change indicates the maximum non-expansion work obtainable from a chemical reaction and determines whether a reaction is spontaneous. In electrochemical cells, it is directly related to the EMF by the equation ?rG = -nFE, where n is the number of moles of electrons transferred and F is the Faraday constant.

Electromotive Force

The electromotive force (EMF) is a measure of the potential difference between two electrodes in an electrochemical cell. It represents the driving force behind the electron flow in a redox reaction and is a key observable in linking electrochemical measurements to thermodynamic quantities.

Transcript

00:01 In this question we have been given a cell which is cadmium cell cadmium solid to cadmium chloride which has the concentration of 0 .1 molar and now the cathode side is having agcl solid to ag solid right so here the cell reaction is given as solid cadmium plus two moles of agcl which is in solid form giving out two moles of ag solid plus cadmium plus two iron and plus two moles of chloride ions and these both of these ions are in solution or we can say in equest form so we can say that the standard emf of this cell the standard emf of this cell at 0 degree celsius and at 25 degrees celsius is given as 0 .6915 volts and 0 .6753 volts respectively right this is the value of emf of this cell at two different temperature conditions now we need to identify the enthalpy change delta h at 25 degree celsius.

01:06 This is what we need to figure out in this particular question.

01:10 So here in this question we are going to use a simple concept or a simple formula that is delta g equals to delta h minus t delta s this is a very typical formula where delta g is the change in gives free energy delta h is the enthalpy change and delta s is the entropy change while t is temperature right so using this equation if we just find out the value of delta g delta s and temperature value is already given to us which is 25 degrees celsius we can easily find out the value of delta h so let us let it as equation number one and let's proceed to uh let's proceed to calculate the value of delta g so delta g would be given by the simple formula which is minus n f e cell n is the moles of electrons involved f is ferrade and e cell is the e cell is the emf of the cell right so here we can say that uh two moles of electrons are being involved so we can write 2 in place of n then ferradi has 96 500 koulaum worth of value and e cell is already given to us which is 0 .673 volts right so here we can say that the resulting value would be 1 30333 jules so this is the value of delta g here and also we can just simply find out the value of delta s which has the simple formula here and the formula of delta s that we use would be equal to nf delta e devalta by delta t at a constant pressure right so n is again the moles of electrons that is that is nothing else but two and then f is 96 000 500 kool -hs while delta e is the change in potentials right and delta t is the change in temperature so we can write it as e2 minus e1 divided by t2 minus t1 so after putting values we can get 2 multiplied by 96 500 koums and multiplied by multiplied by 0 .6753 minus 0 .6915 divided by the temperature values that are 298 minus 273 right and we have written 298 in place of 25 degrees celsius and in place of zero degrees celsius we have written 273 kelvin so we have just converted other celsius temperatures to kelvin temperatures right we have done nothing else now solving this what we can get is we can get the value of delta s that is equal to minus 125 .064 jule kelvin inverse.

03:37 Mold inverse...

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