At 298K , a 1 litre solution containing 10mmol of Cr _ 2 O _ 7 ^ 2...

The balanced half-reaction can be written as follows,

${Cr}_{2}{O}_{7}^{2 -} + 14{H}^{+} + 6{e}^{-} \to 2{Cr}^{3 +} + 7{H}_{2}O$

Using Nernst equation

${E}_{cell} = {E}_{cell}^{o} - \dfrac{0 . 059}{n}log\dfrac{\left[products\right]}{\left[reactants\right]}$

For the above reaction, number of electrons involved n = 6.

$E = 1 . 33 - \dfrac{0 . 059}{6}log\dfrac{{\left(0 . 1\right)}^{2}}{\left({10}^{- 2}\right){\left({10}^{- 3}\right)}^{14}}$

$E = 1 . 33 - \dfrac{0 . 059}{6} \times 42 = 0 . 917$

$E = 917 \times {10}^{- 3}$

$x = 917$.