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CUET · CHEMISTRY · PYQ PAPER 2023

The standard reduction potential for \(\text{Sn}^{4+}/\text{Sn}^{2+}\) is \(+0.15\text{ V}\) and for \(\text{Cr}^{3+}/\text{Cr}\) is \(-0.74\text{ V}\). These two half-cells, coupled in their standard states, are connected to make a cell. The galvanic cell can be correctly represented by :

  1. A \(\text{Sn}^{2+}(\text{aq})|\text{Sn}^{4+}(\text{aq})\) || \(\text{Cr}(\text{s})|\text{Cr}^{3+}(\text{aq})\)
  2. B \(\text{Sn}^{4+}(\text{aq})|\text{Sn}^{2+}(\text{aq})\) || \(\text{Cr}^{3+}(\text{aq})|\text{Cr}(\text{s})\)
  3. C \(C r(s)\left|C r^{3+}(a q) \| S n^{4+}(a q)\right| S n^{2+}(a q)\)
  4. D \(\text{Cr}(\text{s})|\text{Cr}^{3+}(\text{aq})\) || \(\text{Sn}^{2+}(\text{aq})|\text{Sn}^{4+}(\text{aq})\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(C r(s)\left|C r^{3+}(a q) \| S n^{4+}(a q)\right| S n^{2+}(a q)\)

Step-by-step Solution

Detailed explanation

\(E^\circ_{\text{red}}(\text{Cr}^{3+}/\text{Cr}) = -0.74\text{ V}\) is less positive than \(E^\circ_{\text{red}}(\text{Sn}^{4+}/\text{Sn}^{2+}) = +0.15\text{ V}\). Cr will be oxidized (anode), and Sn will be reduced (cathode). Anode:…