1. Write the relevant half-reaction

For dichromate reduction in acidic medium:

$\mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O}$

Given standard reduction potential:

$E^\circ = 1.330\,\text{V}$

2. Use the Nernst equation

At $298\,\text{K}$,

$E = E^\circ - \frac{0.059}{n}\log Q$

Here $n=6$.

For the reaction,

$Q = \frac{[\mathrm{Cr^{3+}}]^2}{[\mathrm{Cr_2O_7^{2-}}][\mathrm{H^+}]^{14}}$

So,

$E = 1.330 - \frac{0.059}{6}\log\left(\frac{[\mathrm{Cr^{3+}}]^2}{[\mathrm{Cr_2O_7^{2-}}][\mathrm{H^+}]^{14}}\right)$

3. Substitute the concentrations

Since the solution volume is $1\,\text{L}$:

• $10\,\text{mmol } \mathrm{Cr_2O_7^{2-}} \Rightarrow [\mathrm{Cr_2O_7^{2-}}] = 0.010\,\text{M}$
• $100\,\text{mmol } \mathrm{Cr^{3+}} \Rightarrow [\mathrm{Cr^{3+}}] = 0.100\,\text{M}$
• $\text{pH}=3 \Rightarrow [\mathrm{H^+}] = 10^{-3}\,\text{M}$

Thus,

$Q = \frac{(0.100)^2}{(0.010)(10^{-3})^{14}}$

Now,

$(0.100)^2 = 10^{-2}$

So,

$Q = \frac{10^{-2}}{10^{-2}\cdot 10^{-42}} = 10^{42}$

Hence,

$\log Q = 42$

4. Calculate the electrode potential

$E = 1.330 - \frac{0.059}{6}\times 42$

Since,

$\frac{42}{6}=7$

therefore,

$E = 1.330 - 0.059\times 7$

$E = 1.330 - 0.413$

$E = 0.917\,\text{V}$

5. Express in the required form

Given:

$E = x \times 10^{-3}\,\text{V}$

Since,

$0.917\,\text{V} = 917 \times 10^{-3}\,\text{V}$

So,

$x = 917$

6. Comparison with stored correct answer

Stored correct answer = $917$

Our derived answer also = $917$

So the answer agrees.