At 298 K, a 1 litre solution containing 10 mmol of and 100 mmol of shows a pH of 3.0.
Given: ;
and
The potential for the half cell reaction is V. The value of is
Given: ;
and
The potential for the half cell reaction is V. The value of is
Show answer and solution
Answer: 917
Solution
The half-reaction is: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Using Nernst equation:
E = E° - (0.059/6) × log([Cr³⁺]²/([Cr₂O₇²⁻][H⁺]¹⁴])
At pH = 3.0, [H⁺] = 10⁻³ M
[Cr³⁺] = 100 mmol/1 L = 0.1 M
[Cr₂O₇²⁻] = 10 mmol/1 L = 0.01 M
Calculate the reaction quotient Q:
Q = [Cr³⁺]²/([Cr₂O₇²⁻][H⁺]¹⁴)
Q = (0.1)²/((0.01)(10⁻³)¹⁴)
Q = 10⁻²/(10⁻² × 10⁻⁴²)
Q = 10⁻²/10⁻⁴⁴ = 10⁴²
log Q = 42
E = 1.330 - (0.059/6) × 42
E = 1.330 - (0.059 × 7)
E = 1.330 - 0.413
E = 0.917 V = 917 × 10⁻³ V
Therefore x = 917.
Using Nernst equation:
E = E° - (0.059/6) × log([Cr³⁺]²/([Cr₂O₇²⁻][H⁺]¹⁴])
At pH = 3.0, [H⁺] = 10⁻³ M
[Cr³⁺] = 100 mmol/1 L = 0.1 M
[Cr₂O₇²⁻] = 10 mmol/1 L = 0.01 M
Calculate the reaction quotient Q:
Q = [Cr³⁺]²/([Cr₂O₇²⁻][H⁺]¹⁴)
Q = (0.1)²/((0.01)(10⁻³)¹⁴)
Q = 10⁻²/(10⁻² × 10⁻⁴²)
Q = 10⁻²/10⁻⁴⁴ = 10⁴²
log Q = 42
E = 1.330 - (0.059/6) × 42
E = 1.330 - (0.059 × 7)
E = 1.330 - 0.413
E = 0.917 V = 917 × 10⁻³ V
Therefore x = 917.
[JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.)]