For the cell Co|Co^{2+}(C_2)||Co^{2+}(C_1)|Co | vedclass

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Question

(B) The cell reaction is $Co(s) + Co^{2+}(C_1) \rightarrow Co^{2+}(C_2) + Co(s)$.The Nernst equation for this concentration cell is $E_{cell} = E_{cel

Vedclass · Accepted Answer

$C_2 < C_1$

Vedclass · Answer

(B) The cell reaction is $Co(s) + Co^{2+}(C_1) \rightarrow Co^{2+}(C_2) + Co(s)$.The Nernst equation for this concentration cell is $E_{cell} = E_{cell}^{\circ} - \frac{0.0591}{n} \log \frac{C_2}{C_1}$.Since $E_{cell}^{\circ} = 0$ for a concentration cell,$E_{cell} = -\frac{0.0591}{n} \log \frac{C_2}{C_1} = \frac{0.0591}{n} \log \frac{C_1}{C_2}$.For the reaction to be spontaneous,$\Delta G$ must be negative,which implies $E_{cell}$ must be positive.$E_{cell} > 0$ when $\log \frac{C_1}{C_2} > 0$,which means $\frac{C_1}{C_2} > 1$,or $C_1 > C_2$.

For the cell $Co|Co^{2+}(C_2)||Co^{2+}(C_1)|Co$,$\Delta G$ is negative if

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