For the following electrochemical cell Sn(s)|Sn2 (aq, 0.0155 M)||Ag (aq, 3.50 M)|Ag(s)

For the following electrochemical cell Sn(s)|Sn2 (aq, 0.0155 M)||Ag (aq, 3.50 M)|Ag(s) write the net cell equation. Phases are optional. Do not include the concentrations. Calculate the following values at 25.0 °C using standard potentials as needed

The Correct Answer and Explanation is :

Electrochemical Cell: Sn(s) | Sn²⁺(aq) || Ag⁺(aq) | Ag(s)

Step 1: Write Half-Cell Reactions

For the Sn electrode (oxidation half-reaction):

[
\text{Sn(s)} \rightarrow \text{Sn}^{2+}(aq) + 2e^-
]

For the Ag electrode (reduction half-reaction):

[
\text{Ag}^+(aq) + e^- \rightarrow \text{Ag(s)}
]

Step 2: Determine the Standard Reduction Potentials

From standard tables:

Sn²⁺/Sn: ( E^\circ = -0.14 \, \text{V} ) (for reduction)
Ag⁺/Ag: ( E^\circ = +0.80 \, \text{V} ) (for reduction)

Since oxidation occurs at the anode and reduction at the cathode, the half-reaction at the anode involves the oxidation of Sn, and at the cathode, Ag⁺ is reduced to Ag.

Step 3: Calculate the Cell Potential (( E^\circ_{\text{cell}} ))

The standard cell potential is calculated using:

[
E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} – E^\circ_{\text{anode}}
]

Substituting values:

[
E^\circ_{\text{cell}} = 0.80 \, \text{V} – (-0.14 \, \text{V}) = 0.80 \, \text{V} + 0.14 \, \text{V} = 0.94 \, \text{V}
]

Thus, the standard cell potential is ( 0.94 \, \text{V} ).

Step 4: Calculate the Nernst Equation

Since the concentrations of ions in the cell are non-standard, we need to use the Nernst equation to calculate the actual cell potential at the given conditions:

[
E = E^\circ_{\text{cell}} – \frac{0.0592}{n} \log Q
]

Where:

( n = 2 ) (the number of electrons transferred),
( Q ) is the reaction quotient, which is given by:

[
Q = \frac{[\text{Sn}^{2+}]{\text{cathode}}}{[\text{Ag}^+]{\text{anode}}}
]

From the problem:

( [\text{Sn}^{2+}]_{\text{cathode}} = 0.0155 \, \text{M} ),
( [\text{Ag}^+]_{\text{anode}} = 3.50 \, \text{M} ).

Thus:

[
Q = \frac{0.0155}{3.50} = 0.00443
]

Now, we can calculate ( E ):

[
E = 0.94 \, \text{V} – \frac{0.0592}{2} \log(0.00443)
]

[
E = 0.94 \, \text{V} – \frac{0.0592}{2} (-2.653)
]

[
E = 0.94 \, \text{V} + 0.0786 \, \text{V}
]

[
E = 1.0186 \, \text{V}
]

So, the actual cell potential is ( E = 1.02 \, \text{V} ) at 25°C.

Step 5: Write the Net Cell Reaction

The overall cell reaction is the sum of the half-reactions:

Anode (oxidation): ( \text{Sn(s)} \rightarrow \text{Sn}^{2+}(aq) + 2e^- )
Cathode (reduction): ( \text{Ag}^+(aq) + e^- \rightarrow \text{Ag(s)} )

To balance the number of electrons, multiply the reduction half-reaction by 2:

[
\text{2Ag}^+(aq) + 2e^- \rightarrow \text{2Ag(s)}
]

Thus, the net reaction is:

[
\text{Sn(s)} + 2\text{Ag}^+(aq) \rightarrow \text{Sn}^{2+}(aq) + 2\text{Ag(s)}
]

Conclusion

The overall cell reaction for the given electrochemical cell is: