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Uploaded ImageWhat 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

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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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