At 298 K, a 1 litre solution containing 10m mol of Cr2O72- and 100m mol Cr3+ shows a pH of 3.0.Given: Cr2O72+ → Cr3+; E0 = 1.330V and 2.303RTF = 0.059 VThe potential for the half cell reaction is x×10-3V. The value of x is___

Cr2O72−+14H++6e−→2Cr+3+7H2OEH.cell =E0−0.0596log⁡Cr+32Cr2O72−H+14=1.33−0.0596log⁡(0.1)2(0.01)10−314= 0.917 V= 9.17 x 10-3 V

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At 298 K, a 1 litre solution containing 10m mol of Cr₂O₇^2- and 100m mol Cr³⁺ shows a pH of 3.0.
Given: Cr₂O₇²⁺ → Cr³⁺; E⁰ = 1.330V and $\frac{2.303 RT}{F}$ = 0.059 V
The potential for the half cell reaction is $x \times 10^{- 3} V$ . The value of x is___

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answer is 917.

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Detailed Solution

$\begin{array}{l} {Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 {Cr}^{+ 3} + 7 H_{2} O \\ E_{H.cell} = E^{0} - \frac{0.059}{6} \log \frac{{[{Cr}^{+ 3}]}^{2}}{[{Cr}_{2} O_{7}^{2 -}] {[H^{+}]}^{14}} \\ = 1.33 - \frac{0.059}{6} \log \frac{(0.1)^{2}}{(0.01) {(10^{- 3})}^{14}} \end{array}$
= 0.917 V
= 9.17 x 10^-3 V