Consider the electrochemical cell shown in th | vedclass

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(C) For a concentration cell, the cell reaction is $M^{+}(c_{cathode}) \rightarrow M^{+}(c_{anode})$.Using the Nernst equation: $E_{cell} = E^{\circ}_

Vedclass · Accepted Answer

If $c_{1}$ is present at the cathode, then $c_{1} > c_{2}$

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(C) For a concentration cell, the cell reaction is $M^{+}(c_{cathode}) \rightarrow M^{+}(c_{anode})$.Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[M^{+}]_{anode}}{[M^{+}]_{cathode}}$.Since $E^{\circ}_{cell} = 0$ for a concentration cell, $E_{cell} = -0.0591 \log \frac{c_{anode}}{c_{cathode}} = 0.0591 \log \frac{c_{cathode}}{c_{anode}}$.For $E_{cell} > 0$, we must have $\log \frac{c_{cathode}}{c_{anode}} > 0$, which implies $c_{cathode} > c_{anode}$.Case $I$: If $c_{1}$ is at the cathode, then $c_{1} > c_{2}$ (Option $C$ is correct).Case $II$: If $c_{1}$ is at the anode, then $c_{2} > c_{1}$ (Option $D$ is incorrect).

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