Question
Find the potentials of the following electrochemical cell:$\mathrm{Cd}\left|\mathrm{Cd}^{2+}, M=0.10 \| \mathrm{Ni}^{2+}, M=0.50\right| \mathrm{Ni}$
Step 1
The half-reactions are as follows: \begin{align*} \text{Cd} &\rightarrow \text{Cd}^{2+} + 2e^- \\ \text{Ni}^{2+} + 2e^- &\rightarrow \text{Ni} \end{align*} Show more…
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The voltaic cell $$\mathrm{Cd}(s)\left|\mathrm{Cd}^{2+}(a q) \| \mathrm{Ni}^{2+}(1.0 M)\right| \mathrm{Ni}(s)$$ has a cell potential of $0.290 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. What is the concentration of cadmium ion? $\left(E_{\text {cell }}^{\circ}=0.170 \mathrm{~V} .\right)$ .
An electrochemical cell consists of a silver metal electrode immersed in a solution with $\left[\mathrm{Ag}^{+}\right]=1.0 M$ separated by a porous disk from a copper metal electrode. If the copper electrode is placed in a solution of 5.0$M \mathrm{NH}_{3}$ that is also 0.010$M$ in $\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+},$ what is the cell potential at $25^{\circ} \mathrm{C} ?$ $$\mathrm{CdS}(s)+2 \mathrm{e}^{-} \rightarrow \mathrm{Cd}(s)+\mathrm{S}^{2-}(a q) \qquad \quad \mathscr{E}^{\circ}=-1.21 \mathrm{V}$$ $$\mathrm{Cd}^{2+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Cd}(s) \quad\quad\quad \quad\quad\quad\quad\mathscr{E}^{\circ}=-0.402 \mathrm{V}$$
$ \quad$ The voltaic cell $$ \operatorname{Cd}(s)\left|\mathrm{Cd}^{2+}(a q) \| \mathrm{Ni}^{2+}(1.0 M)\right| \mathrm{Ni}(s) $$ has a cell potential of $0.240 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. What is the concentration of cadmium ion? $\left(E_{\text {cell }}^{\circ}=0.170 \mathrm{~V} .\right)$
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