Consider the following half cell reaction: Cr₂O₇²⁻ (aq) + 6e⁻ + 14H⁺ (aq) → 2Cr³⁺ (aq) + 7H₂O (l) The reaction was conducted with the ratio of [Cr³⁺]² / [Cr₂O₇²⁻] = 10⁻⁶. The pH value at which the EMF of the half cell will become zero is _____. (nearest integer value)
Question: Consider the following half-cell reaction: C r 2 O 7 2 − ( a q ) + 6 e − + 14 H + ( a q ) ⟶ 2 C r 3 + ( a q ) + 7 H 2 O ( l ) The reaction was conducted with the ratio [ C r 3 + ] 2 [ C r 2 O 7 2 − ] = 10 − 6 \dfrac{[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) Short Solution (Text): Step 1: Nernst equation (at 298 K) For the half reaction, E = E ∘ − 0.05916 n log 10 Q where n = 6 n=6 and Q = [ C r 3 + ] 2 [ C r 2 O 7 2 − ] [ H + ] 14 Q=\dfrac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^+]^{14}} Set E = 0 E=0 (given) ⇒ E ∘ = 0.05916 6 log 10 Q Step 2: insert Q Q and given ratio Let R = [ C r 3 + ] 2 [ C r 2 O 7 2 − ] = 10 − 6 R=\dfrac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}]}=10^{-6} . Then Q = R [ H + ] 14 ⇒ log 10 Q = log 10 R − 14 log 10 [ H + ] . Step 3: solve for log 10 [ H + ] \log_{10}[H^+] Using the standard potential E ∘ E^\circ for C r 2 O 7 2 − / C r 3 + \math...