Question:
What would be the cell potential (E) at 25°C for a voltaic cell consisting of Zn/Cu. (Zn | Zn2+ || Cu2+ | Cu) where the concentration
of CuSO4 = 0.05 M and the concentration of ZnS04 = 2.0 M
Redox reaction:
cu2*(ag) + 2e= > Cu (s)
+ -
Zn (s) > Zn? (ag) + 2e
Standard cell potential E°cell = 1.10 V
E=FE° —(RT/nF)InQ
E° = standard cell potential
R=8.31)/mol*K T= Kelvin Temperature
F=9.64110% C/mol n= # electrons transfered
+17.7V
+1.15V
+1.10V
+1.34V
This is AI generated Response.
Asked by: Test User 100137
Created at: 2025-05-07 04:02:08
Asked by: Test User 100137
Created at: 2025-05-07 04:02:08
Answer:
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Let's solve the problem step by step using the **Nernst equation**:
\[
E = E^\circ - \frac{RT}{nF} \ln Q
\]
Given:
- \( E^\circ_{cell} = 1.10 \) V
- \( [\text{Cu}^{2+}] = 0.05 \) M
- \( [\text{Zn}^{2+}] = 2.0 \) M
- \( T = 25^\circ C = 298 \) K
- \( R = 8.31 \) J/mol·K
- \( F = 9.64 \times 10^4 \) C/mol
- \( n = 2 \) electrons
**Step 1: Write the overall cell reaction**
\[
\text{Zn (s)} + \text{Cu}^{2+} (aq) \rightarrow \text{Zn}^{2+} (aq) + \text{Cu (s)}
\]
**Step 2: Write the reaction quotient, \( Q \)**
\[
Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}
= \frac{2.0}{0.05} = 40
\]
**Step 3: Plug values into the Nernst equation**
\[
E = 1.10 - \frac{(8.31)(298)}{(2)(9.64 \times 10^4)} \ln(40)
\]
Calculate the denominator:
\[
2 \times 9.64 \times 10^4 = 1.928 \times 10^5
\]
Calculate the numerator:
\[
8.31 \times 298 = 2477.38
\]
Calculate:
\[
\frac{2477.38}{1.928 \times 10^5} \approx 0.01285
\]
Calculate \( \ln(40) \):
\[
\ln(40) \approx 3.689
\]
Now multiply:
\[
0.01285 \times 3.689 \approx 0.0474
\]
Now subtract from 1.10:
\[
E = 1.10 - 0.0474 = 1.0526 \approx \boxed{1.05~\text{V}}
\]
Since the closest answer choice is **+1.05 V**, but among the given options, the closest is **+1.10 V**.
**However, based on the math, the actual value is slightly less than 1.10 V, but since +1.10 V is the closest, that is the answer you should select from the choices provided.**
**Final Answer:**
\[
\boxed{+1.10~\text{V}}
\]
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