Question

For the galvanic cell,
Zn(s) + Cu\(^{2+}\) (0.02 M) \(\rightarrow\) Zn\(^{2+}\) (0.04 M) + Cu(s),
E\(_{cell}\) = __________ \(\times\) 10\(^{-2}\) V. (Nearest integer)
[Use : E\(^0_{Cu/Cu^{2+}}\) = -0.34 V, E\(^0_{Zn/Zn^{2+}}\) = +0.76 V, \(\frac{2.303 RT}{F} = 0.059V\)]

Hint:

Be very careful with the sign conventions for electrode potentials. The question might give oxidation potentials (like E\(^0_{M/M^{n+}}\)). Remember that the standard reduction potential E\(^0_{M^{n+}/M}\) has the opposite sign. The formula \(E^0_{cell} = E^0_{cathode} - E^0_{anode}\) always uses standard \textbf{reduction} potentials.

Answer 1

Correct Answer: 109

Step 1: Understanding the Question
We need to calculate the cell potential (E\(_{cell}\)) for a galvanic cell under non-standard conditions using the Nernst equation.
Step 2: Key Formula or Approach
The Nernst equation for a cell is:
\[ E_{cell} = E^0_{cell} - \frac{0.059}{n} \log Q \] where \(E^0_{cell}\) is the standard cell potential, \(n\) is the number of electrons transferred, and \(Q\) is the reaction quotient.
Step 3: Detailed Calculation
Calculate the Standard Cell Potential (\(E^0_{cell}\)):
The overall reaction is Zn + Cu\(^{2+}\) \(\rightarrow\) Zn\(^{2+}\) + Cu.
Oxidation (Anode): Zn \(\rightarrow\) Zn\(^{2+}\) + 2e\(^-\)
Reduction (Cathode): Cu\(^{2+}\) + 2e\(^-\) \(\rightarrow\) Cu
We need the standard reduction potentials:
E\(^0_{Zn^{2+}/Zn}\): Given E\(^0_{Zn/Zn^{2+}}\) = +0.76 V (standard oxidation potential), so E\(^0_{Zn^{2+}/Zn}\) = -0.76 V (standard reduction potential).
E\(^0_{Cu^{2+}/Cu}\): Given E\(^0_{Cu/Cu^{2+}}\) = -0.34 V (standard oxidation potential), so E\(^0_{Cu^{2+}/Cu}\) = +0.34 V (standard reduction potential).
Now, calculate \(E^0_{cell}\):
\[ E^0_{cell} = E^0_{cathode} - E^0_{anode} = E^0_{Cu^{2+}/Cu} - E^0_{Zn^{2+}/Zn} \] \[ E^0_{cell} = (+0.34 \text{ V}) - (-0.76 \text{ V}) = 1.10 \text{ V} \] Determine n and Q:
From the half-reactions, the number of electrons transferred, \(n\), is 2.
The reaction quotient, \(Q\), is:
\[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{0.04 \text{ M}}{0.02 \text{ M}} = 2 \] Apply the Nernst Equation:
\[ E_{cell} = 1.10 - \frac{0.059}{2} \log(2) \] Using log(2) \(\approx\) 0.301:
\[ E_{cell} = 1.10 - (0.0295 \times 0.301) \] \[ E_{cell} = 1.10 - 0.0088795 \approx 1.0911 \text{ V} \] Express the answer in the required format:
\[ E_{cell} = 1.0911 \text{ V} = 109.11 \times 10^{-2} \text{ V} \] Step 4: Final Answer
Rounding to the nearest integer, the value is 109.