Consider the following half cell reaction: $$Cr_2O_7^{2-}(aq) + 6e^- + 14H^+(aq) \to 2Cr^{3+}(aq) + 7H_2O(l)$$. The reaction was conducted with the ratio of $$\frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}]} = 10^{-6}$$. The pH value at which the EMF of the half cell will become zero is _____ . (nearest integer value)  [Given : standard half cell reduction potential $$E^0_{Cr_{2}O^{2-}_{7},H^{+}/Cr^{3+}} = 1.33$$ V, $$\frac{2.303RT}{F} = 0.059$$ V]

The given reduction half-cell is
$$Cr_2O_7^{2-}(aq)+14H^+(aq)+6e^- \rightarrow 2Cr^{3+}(aq)+7H_2O(l)$$

Step 1: Write the Nernst equation
For a general reduction reaction, $$E = E^{0}-\dfrac{0.059}{n}\log_{10}Q$$ where
$$n =$$ number of electrons transferred (= 6 here) and
$$Q =$$ reaction quotient.

Step 2: Express the reaction quotient $$Q$$
Only aqueous species appear in $$Q$$, hence
$$Q = \dfrac{[Cr^{3+}]^{2}}{[Cr_2O_7^{2-}][H^+]^{14}}$$

Step 3: Insert the given concentration ratio
We are told $$\dfrac{[Cr^{3+}]^{2}}{[Cr_2O_7^{2-}]} = 10^{-6}$$.
Therefore
$$Q = \dfrac{10^{-6}}{[H^+]^{14}}$$

Step 4: Set the electrode potential to zero
The emf becomes zero when $$E = 0$$, so
$$0 = E^{0}-\dfrac{0.059}{6}\log Q$$
$$\Rightarrow \log Q = \dfrac{6E^{0}}{0.059}$$

Step 5: Calculate $$\log Q$$
Given $$E^{0} = 1.33\,$$V,
$$\log Q = \dfrac{6 \times 1.33}{0.059} = \dfrac{7.98}{0.059} \approx 135.25$$

Step 6: Relate $$\log Q$$ to $$[H^+]$$
From Step 3,
$$Q = \dfrac{10^{-6}}{[H^+]^{14}}$$
Take $$\log_{10}$$ of both sides:
$$\log Q = \log(10^{-6}) - 14\log[H^+]$$
$$\Rightarrow 135.25 = -6 - 14\log[H^+]$$

Step 7: Solve for $$\log[H^+]$$
$$-14\log[H^+] = 135.25 + 6 = 141.25$$
$$\log[H^+] = -\dfrac{141.25}{14} \approx -10.09$$

Step 8: Obtain the pH
$$[H^+] = 10^{-10.09}$$
$$pH = -\log[H^+] = 10.09 \approx 10$$

Therefore, the emf of the given half-cell becomes zero at pH = 10.