At 298 K, a 1 litre solution containing 10 m mol of Cr 2 O 7 2- and 100 m mol of Cr 3+ shows a pH of 3.0. Given: Cr 2 O 72- arrow Cr 3+ ; E °=1.330 V and (2.303 RT / F )=0.059 V The potential for the half cell reaction is x × 10-3 V. The value of x is

Question

At 298 K, a 1 litre solution containing 10 m mol of Cr 2 O 7 2- and 100 m mol of Cr 3+ shows a pH of 3.0. Given: Cr 2 O 72- arrow Cr 3+ ; E °=1.330 V and (2.303 RT / F )=0.059 V The potential for the half cell reaction is x × 10-3 V. The value of x is  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Cr 2 O 72-+14 H ++6 e - arrow 2 Cr 3++7 H 2 O E =1.33-(0.059/6) log ((0.1)2/(10-2)(10-3)14) E =1.33-(0.059/6) × 42=0.917 E =917 × 10-3 x =917

Q. At 298K, a 1 litre solution containing 10mmol of Cr2​O7​2− and 100mmol of Cr3+ shows a pH of 3.0. Given: Cr2​O72−​→Cr3+;E∘=1.330V and F2.303RT​=0.059V The potential for the half cell reaction is x×10−3V. The value of x is ___

Cr2​O72−​+14H++6e−→2Cr3++7H2​O E=1.33−60.059​log(10−2)(10−3)14(0.1)2​ E=1.33−60.059​×42=0.917 E=917×10−3 x=917

$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$
$E = 1.33 - \frac{0.059}{6} lo g \frac{( 0.1 ) ^{2}}{( 1 0 ^{- 2} ) ( 1 0 ^{- 3} ) ^{14}}$
$E = 1.33 - \frac{0.059}{6} \times 42 = 0.917$
$E = 917 \times 1 0^{- 3}$
$x = 917$