Question

Calculate the EMF between copper and silver Ag+e-E=0.89vCu=E=0.34v​

Answer 1

Answer:

Depending on the $E^\circ$ value of $\rm Ag^{+} + e^{-} \to Ag\; (s)$, the cell potential would be:

• $0.55\; \rm V$, using data from this particular question; or
• approximately $0.46\; \rm V$, using data from the CRC handbooks.

Explanation:

In this galvanic cell, the following two reactions are going on:

• The conversion between $\rm Ag\; (s)$ and $\rm Ag^{+}$ ions, $\rm Ag^{+} + e^{-} \rightleftharpoons Ag\; (s)$, and
• The conversion between $\rm Cu\; (s)$ and $\rm Cu^{2+}$ ions, $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$.

Note that the standard reduction potential of $\rm Ag^{+}$ ions to $\rm Ag\; (s)$ is higher than that of $\rm Cu^{2+}$ ions to $\rm Cu\; (s)$. Alternatively, consider the fact that in the metal activity series, copper is more reactive than silver. Either way, the reaction is this cell will be spontaneous (and will generate a positive EMF) only if $\rm Ag^{+}$ ions are reduced while $\rm Cu\; (s)$ is oxidized.

Therefore:

• The reduction reaction at the cathode will be: $\rm Ag^{+} + e^{-} \to Ag\; (s)$. The standard cell potential of this reaction (according to this question) is $E(\text{cathode}) = 0.89\; \rm V$. According to the 2012 CRC handbook, that value will be approximately $0.79\; \rm V$.

• The oxidation at the anode will be: $\rm Cu\; (s) \to \rm Cu^{2+} + 2\, e^{-}$. According to this question, this reaction in the opposite direction ($\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$) has an electrode potential of $0.34\; \rm V$. When that reaction is inverted, the electrode potential will also be inverted. Therefore, $E(\text{anode}) = -0.34\; \rm V$.

The cell potential is the sum of the electrode potentials at the cathode and at the anode:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &= 0.89 \; \rm V + (-0.34\; \rm V) = 0.55\; \rm V\end{aligned}$.

Using data from the 1985 and 2012 CRC Handbook:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &\approx 0.7996 \; \rm V + (-0.337\; \rm V) \approx 0.46\; \rm V\end{aligned}$.