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

Q. The electrode with reaction :

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

can be represented as:

Q. The standard electrode potential for the following reaction is

- 0.57 V

T e O_{3}^{2 -} (a q, 1 M) + 3 H_{2} O (l) + 4 e^{-} \to T e (s) + 6 O H^{-} (a q)

Q. The half cell involving the reaction

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

can be represented as :

Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox Titration. Some half – cell reactions and their standard potential are given below:
$M n O_{4}^{-} (a q) + 8 H^{+} (a q) + 5 e^{-} \to M n^{2 +} (a q) + 4 H_{2} O (I); E^{\circ} = 1.51 V$
$C r_{2} O_{7}^{2 -} (a q) + 14 H^{+} (a q) + 6 e^{-} \to 2 C r^{3 +} (a q) + 7 H_{2} O (I); E^{\circ} = 1.38 V$
$F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); E^{\circ} = 0.77 V$
$C l_{2} (g) + 2 e^{-} \to 2 C l^{-} (a q); E^{\circ} = 1.40 V$

Identify the only incorrect statement regarding the quantitative estimation of aqueous $F e (N O_{3})_{2}$ .

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Cr2O2−7(aq,1M)+14H+(aq)+6e−→2Cr3+(aq,1M)+7H2O. The standard electrode potential for the following reaction is + 1.33V. what is the potential at pH=2.0 ?

The correct option is D + 1.05 VpH=2.0⇒(H+)=10−2MQC=[Cr3+]2[Cr2O2+7][H+]14QC=121×(10−2)14=1028ECr2O2−7/Cr3+=E0Cr2O2−7/Cr3++0.06nlog1Qc=1.33+0.066log 10−28=1.05

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