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

+ 1.82 V

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+ 1.99 V

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+ 1.61 V

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+ 1.05 V

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Solution

The correct option is D + 1.05 V
$p H = 2.0 \Rightarrow (H^{+}) = 10^{- 2} M Q_{C} = \frac{[C r^{3 +}]^{2}}{[C r_{2} O_{7}^{2 +}] [H^{+}]^{14}} Q_{C} = \frac{1^{2}}{1 \times (10^{- 2})^{14}} = 10^{28} E_{C r_{2} O_{7}^{2 -} / C r^{3 +}} = E_{C r_{2} O_{7}^{2 -} / C r^{3 +}}^{0} + \frac{0.06}{n} l o g \frac{1}{Q c} = 1.33 + \frac{0.06}{6} l o g 10^{- 28} = 1.05$

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Similar questions

C r_{2} O_{7}^{2 -} (a q, 1 M) + 14 H^{+} (a q) + 6 e^{-} \to 2 C r^{3 +} (a q, 1 M) + 7 H_{2} O

p H = 2.0

C r_{2} O_{7}^{2 -} (a q) + 14 H^{\oplus} (a q) + 6 e^{-} \to 2 C r^{3 +} (a q) + 7 H_{2} O

C r_{2} O_{7}^{2 -} (a q) + 14 H^{+} (a q) + 6 e \to 2 C r^{3 +} (a q) + 7 H_{2} O

M n_{4}^{-} (a q) + 8 H^{+} (a q) + 5 e^{-} \to M n^{2 +} (a q) + 4 H_{2} O (l); E^{\circ} = 1.51 V

C r_{2} O_{7}^{2 -} (a q) + 14 H^{+} + 6 e^{-} \to 2 C r^{3} + (a q) + 7 H_{2} O (l); E^{\circ} = 138 V

F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); E^{\circ} = 0.77 V

C l_{2} (g) + 2 e^{-} \to 2 C l^{-} (a q); E^{\circ} = 1.40 V

F e (N O_{3})_{2}

M n O_{4}^{-} (a q .) + 8 H^{+} (a q .) + 5 e^{-} \to M n^{2 +} (a q .) + 4 H_{2} O (l); E^{o} = 1.51 V

C r_{2} O_{7}^{2 -} (a q .) + 14 H^{+} (a q .) + 6 e^{-} \to 2 C r^{3 +} (a q .) + 7 H_{2} O (l); E^{o} = 1.38 V

F e^{3 +} (a q .) + e^{-} \to F e^{2 +} (a q .); E^{o} = 0.77 V

C l_{2} (g) + 2 e^{-} \to 2 C l^{-} (a q .); E^{o} = 1.40 V

F e (N O_{3})_{2}

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