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Answer (Detailed Solution Below)

Option 3 : $-0.85$ V / $-0.85$ V

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Detailed Solution

सही उत्तर $-0.85$ V / $-0.85$ V है।

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PrepSnapprepsnap.in • Part -A

Question:

For the electrode reaction $\mathrm{Zn^{2+} + 2e^- \rightarrow Zn}$ with $E^\circ = -0.76$ V, the electrode potential when $[\mathrm{Zn^{2+}}] = 10^{-3}$ M at $298$ K is / इलेक्ट्रोड अभिक्रिया $\mathrm{Zn^{2+} + 2e^- \rightarrow Zn}$ हेतु $E^\circ = -0.76$ V है, $298$ K पर जब $[\mathrm{Zn^{2+}}] = 10^{-3}$ M हो तो इलेक्ट्रोड विभव है

1.
$-0.67$ V / $-0.67$ V
2.
$-0.73$ V / $-0.73$ V
3.
$-0.85$ V / $-0.85$ V
4.
$-0.79$ V / $-0.79$ V

Correct Answer: $-0.85$ V / $-0.85$ V

$E = E^\circ + (0.0591/2)\log[\mathrm{Zn^{2+}}] = -0.76 + 0.02955 \times (-3) = -0.85$ V. Lowering the ion concentration makes the reduction potential more negative. / $E = E^\circ + (0.0591/2)\log[\mathrm{Zn^{2+}}] = -0.76 + 0.02955 \times (-3) = -0.85$ V। आयन सांद्रता घटाने से अपचयन विभव अधिक ऋणात्मक होता है।

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