One half-cell in a voltaic cell is constructed from a silver wire dipped into a AgNO 3 solution

One half-cell in a voltaic cell is constructed from a silver...

Key Concepts

Nernst Equation

The Nernst equation is a fundamental formula in electrochemistry that relates the electrode potential of a half-cell to its standard reduction potential, temperature, number of electrons transferred, and the reaction quotient. It allows calculation of cell potentials under non-standard conditions by accounting for variations in ion concentration or pressure.

Reaction Quotient

The reaction quotient (Q) is a dimensionless number that expresses the ratio of the concentrations or activities of the products to those of the reactants, each raised to a power equal to its stoichiometric coefficient. In the context of electrochemistry, Q is used in the Nernst equation to adjust the potential from standard conditions to the actual conditions within the cell.

Voltaic Cell

A voltaic cell, also known as a galvanic cell, is an electrochemical device that converts chemical energy from spontaneous redox reactions into electrical energy. It comprises two half-cells, each with its own electrode and electrolyte, connected by a conductive path and a salt bridge that allows ion flow to balance charge during the reaction.

Standard Reduction Potentials

Standard reduction potentials are tabulated values that indicate the tendency of a chemical species to gain electrons under standard conditions. These potentials are critical in determining which half-reaction occurs as oxidation and which as reduction in an electrochemical cell, thus enabling prediction and calculation of the cell's overall voltage.

Transcript

00:01 For this problem, we are given information about the voltage of an entire galvanic cell and are asked to work backwards in order to find the concentration of silver cation in solution in one of the half cells.

00:19 Now, before i begin, i would just like to say that the values, again, that i found for reduction potentials might be different than the ones that you have in your book.

00:31 However the math should be the same and the answers are probably pretty close to each other so just make sure you're using the values that the book gives you instead of the ones that i have in this video okay the other thing that i would like to point out real quick is that i've included the important equations that you're going to use in order to solve this is the nursed equation at stp where n is the number of electrons being transferred into reaction and q is the reaction quotient and then this equation down here simply states that the voltage of an entire voltaic cell is the difference between e plus and e minus where e plus is your more positive reduction potential and e minus is your more negative reduction potential okay, so to begin we first need to identify which species is being oxidized and which species is being reduced.

01:45 And we can do that by looking at reduction potentials.

01:51 I've gone ahead and written out the overall reaction at the top here.

01:58 You can assign oxidation states there if you want, but you don't have to for this part because we have our little table of reduction potentials right here.

02:09 So we know based on these values that ag plus that this half cell, whoops, must be e plus because it is more positive than this value, so that must be e minus for zinc and e plus for ag.

02:30 Okay.

02:31 So basically that's important because we need to plug in these values down here after we use the nernst equation to solve for e minus.

02:40 So we're given the concentration of zinc, but we're not given the concentration of silver, we're asked to find the concentration of silver, but we are given the overall cell voltage.

02:53 So we can use this equation down here in order to solve for e plus, and then we can go back and use the nernst equation to solve for our concentration, excuse me.

03:11 Okay, so let's start by solving e minus.

03:19 E minus is going to be equal to negative .7618 volts minus 0 .05916 volts over 2 because we're transferring two electrons in the half reaction multiplied by log times the reaction quotient, which is just going to be 1 .0 molar.

03:53 Actually, i wrote that wrong, should be 1 over 1 .0 molar, which is just going to equal 1.

04:02 And whenever you have a logarithm with a base of 10, and then a 1, a value of 1 here in parentheses, you know that that's equal to 0.

04:15 So this whole term right here is going to go to 0, which means that e minus is just going to be equal to negative 0 .7618 volts.

04:34 So now we can plug this value in this equation up here and solve for e plus.

04:43 So we get 1 .48 volts is equal to e plus minus negative 0 .7618 volts.

04:57 And we find that e plus has a value of .7182 volts...

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