SATHEE: Redox Reactions And Electrochemistry Ques 735

Redox Reactions And Electrochemistry Ques 735

Question: The standard electrode potential for the following reaction is +1.33V. What is the potential at $p H = 2.0$ - $C r_{2} O_{7}^{2 -} (a q .1 M) + 14, H^{+} (a q) + 6 e^{-} \overset{}{\to}$ $2 C r^{3 +} (a q ., 1, M) + 7 H_{2} O (ℓ)$

Options:

A) +1.820 V

B) +1.990 V

C) +1.608 V

D) +1.0542 V

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Answer:

Correct Answer: D

Solution:

[d] $E_{C r_{2} O_{7}^{2 -} {| C r .}^{3 +}} = E {^{\circ}}_{C r_{2} O_{7}^{2 -} {| C r .}^{3 +}}, - \frac{0.0591}{6}$ $\times (\log \frac{{[C r^{3 +}]}^{2}}{[C r_{2} O_{7}^{2 -}]} \times \frac{1}{{[H^{+}]}^{14}})$

$E_{C r_{2} O_{7}^{2 -} {| C r .}^{3 +}} = 1.33 - \frac{0.0591}{6} \log \frac{1}{{(0.01)}^{14}}$

$\Rightarrow$ $1.0542, V$

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Redox Reactions And Electrochemistry Ques 735

Question: The standard electrode potential for the following reaction is +1.33V. What is the potential at pH=2.0 - Cr2O72−(aq.1M)+14,H+(aq)+6e−→ 2Cr3+(aq.,1,M)+7H2O(ℓ)

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Question: The standard electrode potential for the following reaction is +1.33V. What is the potential at $p H = 2.0$ - $C r_{2} O_{7}^{2 -} (a q .1 M) + 14, H^{+} (a q) + 6 e^{-} \overset{}{\to}$ $2 C r^{3 +} (a q ., 1, M) + 7 H_{2} O (ℓ)$