Question

Consider the redox reaction $Mg_{(s)}+2~Ag_{(aq)}^{+}\rightarrow Mg_{(aq)}^{2+}+2~Ag_{(s)}$. The expression of cell emf at 298 K is:

• A. $E=E^{\circ}+\frac{0.059}{2}log\frac{[Mg^{2+}]}{[Ag^{+}]^{2}}$
• B. $E=E^{\circ}+\frac{0.059}{2}log\frac{[Ag^{+}]^{2}}{[Mg^{2+}]}$
• C. $E=E^{\circ}-\frac{0.059}{2}log\frac{[Mg^{2+}]}{[Ag]}$
• D. $E=E^{\circ}-\frac{0.059}{2}log\frac{[Mg^{2+}]^{2}}{[Ag^{+}]}$

Hint:

If you use a "$+$" sign before the log in Nernst equation, put the Reactants on top and Products on bottom.

Answer 1

The Correct Option is B

Step 1: Concept The Nernst Equation relates the cell potential to the reaction quotient ($Q$): $E = E^{\circ} - \frac{0.059}{n} log Q$.

Step 2: Meaning For the reaction $Mg_{(s)} + 2Ag^{+} \rightarrow Mg^{2+} + 2Ag_{(s)}$, the number of electrons transferred ($n$) is 2. The reaction quotient $Q = \frac{[Mg^{2+}]}{[Ag^{+}]^{2}}$ (pure solids are excluded).

Step 3: Analysis Substituting $Q$ and $n=2$ into the Nernst equation: $E = E^{\circ} - \frac{0.059}{2} log \frac{[Mg^{2+}]}{[Ag^{+}]^{2}}$. This can be rewritten by inverting the log term and changing the sign: $E = E^{\circ} + \frac{0.059}{2} log \frac{[Ag^{+}]^{2}}{[Mg^{2+}]}$.

Step 4: Conclusion The rewritten expression matches option (B). Final Answer: (B)