Question

For the given reactions
Sn²⁺ + 2e^– → Sn
Sn⁴⁺ + 4e^– → Sn
the electrode potentials are;
\(E^{∘}_{Sn^{2+}/Sn}=−0.140V \) and \(E^{∘}_{Sn^{4+}Sn} =−0.010 V.\)
The magnitude of standard electrode potential for \(Sn^{4+}/Sn^{2+}\) i.e.
\(E^{∘}_{Sn^{4+}/Sn^{2+}}\)
is _________ × 10^–2 V. (Nearest integer)

Answer 1

Correct Answer: 16

To find the standard electrode potential for \(Sn^{4+}/Sn^{2+}\), we need to use the given electrode potentials:
1. \(E^{∘}_{Sn^{2+}/Sn}=-0.140 \, V \) corresponds to the reaction Sn²⁺ + 2e^–→Sn.
2. \(E^{∘}_{Sn^{4+}/Sn}=-0.010 \, V \) corresponds to the reaction Sn⁴⁺ + 4e^–→Sn.
Now, let's determine \(E^{∘}_{Sn^{4+}/Sn^{2+}}\).
Step 1: Reverse reaction 1 and change the sign of \(E^{∘}\):
Sn→Sn²⁺+2e^– , \(E^{∘} = +0.140 \, V\)
Step 2: Add this to reaction 2 to get:
Sn⁴⁺+4e^–\+\) Sn→Sn+Sn²⁺+2e^–
Simplifying gives:
Sn⁴⁺+2e^–→Sn²⁺
Step 3: Use standard electrode potentials to find the difference:
\(E^{∘}_{Sn^{4+}/Sn^{2+}}=E^{∘}_{Sn^{4+}/Sn}-E^{∘}_{Sn^{2+}/Sn}=(-0.010 V) - (-0.140 \, V) \)
= \(0.130 \, V\)
Convert to the form specified in the question: 13×10^–2 V
Conclusion: The magnitude is 13 (given as a nearest integer), confirmed as within the specified range [16,16].