At 298K, 1 litre solution containing 10m mol of Cr2O7−2 and 100 m mol Cr+3 shows a PH of 3.0. Given Cr2O7−2→Cr+3,E0=1.330V & 2.303RTF=0.059V The potential for the half cell reaction is x×10−3V. The value of x is……..

Cr2O7−2+14H++6e−→2Cr+3+7H2OEH.cell=E0−0.0596log[Cr+3]2[Cr2O72−][H+]14=1.33−0.0596log(0.1)2(0.01)(10−3)14=0.917V=917×10−3V

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At 298K, 1 litre solution containing 10m mol of $C r_{2} O_{7}^{- 2}$ and 100 m mol $C r^{+ 3}$ shows a $P^{H}$ of 3.0. Given $C r_{2} O_{7}^{- 2} \to C r^{+ 3}, E^{0} = 1.330 V$ $& \frac{2.303 R T}{F} = 0.059 V$ 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

$C r_{2} O_{7}^{- 2} + 14 H^{+} + 6 e^{-} \to 2 C r^{+ 3} + 7 H_{2} O$
$E_{H . c e l l} = E^{0} - \frac{0.059}{6} \log \frac{{[C r^{+ 3}]}^{2}}{[C r_{2} O_{7}^{2 -}] {[H^{+}]}^{14}}$
$= 1.33 - \frac{0.059}{6} \log \frac{{(0.1)}^{2}}{(0.01) {(10^{- 3})}^{14}}$
$= 0.917 V = 917 \times 10^{- 3} V$