Question:
A student builds a galvanic cell utilizing the reaction
\[
Ni(s)+2Ag^+(aq)\rightarrow Ni^{2+}(aq)+2Ag(s)
\]
\[
E^\circ=1.05V
\]
At \(25^\circ C\), what could the student do for the cell to generate a potential greater than that of the initial standard cell?
A student builds a galvanic cell utilizing the reaction
\[
Ni(s)+2Ag^+(aq)\rightarrow Ni^{2+}(aq)+2Ag(s)
\]
\[
E^\circ=1.05V
\]
At \(25^\circ C\), what could the student do for the cell to generate a potential greater than that of the initial standard cell?
Show Hint
Activities of pure solids are taken as unity; hence changing electrode size does not affect cell emf.
Updated On: Jun 18, 2026
- Increase the concentration of \(Ag^+(aq)\)
- Increase the size of the \(Ni(s)\) electrode
- Decrease the size of the \(Ag(s)\) electrode
- Increase the pressure
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
According to the Nernst equation,
\[
E=E^\circ-\frac{0.0591}{n}\log Q
\]
where \(Q\) is the reaction quotient.
Step 1: Write the reaction quotient.
\[ Q= \frac{[Ni^{2+}]}{[Ag^+]^2} \]
Step 2: Determine how to increase cell potential.
To increase \(E\), \[ Q \] must decrease. Increasing \[ [Ag^+] \] decreases \(Q\). Hence \[ E>E^\circ \]
Step 3: Check other options.
Changing size of solid electrodes does not affect \(Q\). Pressure has no effect because gases are absent. Therefore, \[ \boxed{\text{Increase }[Ag^+]} \]
Step 1: Write the reaction quotient.
\[ Q= \frac{[Ni^{2+}]}{[Ag^+]^2} \]
Step 2: Determine how to increase cell potential.
To increase \(E\), \[ Q \] must decrease. Increasing \[ [Ag^+] \] decreases \(Q\). Hence \[ E>E^\circ \]
Step 3: Check other options.
Changing size of solid electrodes does not affect \(Q\). Pressure has no effect because gases are absent. Therefore, \[ \boxed{\text{Increase }[Ag^+]} \]
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