Question

Find Ecell for an electrochemical cell based on the following reaction with [MnO4−]=1.20M, [H+]=1.50M, and [Ag+]=0.0100M. E∘cell for the reaction is +0.88V. MnO4−(aq)+4H+(aq)+3Ag(s)→MnO2(s)+2H2O(l)+3Ag+(aq)

Answer 1

Using the Nernst equation with the given concentrations, the calculated Ecell for the electrochemical cell reaction is 1.322V.

Step-by-step explanation:

To find the Ecell for the given electrochemical cell, we will use the Nernst equation:

Ecell = E°cell - (RT/nF)lnQ

Where:
E°cell is the standard cell potential
R is the universal gas constant (8.314 J/mol·K)
T is the temperature in Kelvin (298 K at room temperature)
n is the number of moles of electrons transferred
F is the Faraday's constant (96485 C/mol)
Q is the reaction quotient

The reaction quotient, Q, can be calculated using the given concentrations for MnO4-, H+, and Ag+.

Q = ([Ag+]^3) / ([MnO4-][H+]^4)

By plugging in the values:
Q = (0.0100^3) / (1.20 * 1.50^4) = 1.85185 × 10^-8

Since the number of electrons transferred (n) is 3 (as 3 Ag atoms lose an electron each), and assuming the reaction is taking place at room temperature (25°C, which is 298K),

Ecell = E°cell - (RT/nF)lnQ
= 0.88V - (8.314 * 298 / (3 * 96485))ln(1.85185 × 10^-8)
= 0.88V - 0.0257V × ln(1.85185 × 10^-8)
= 0.88V - 0.0257V × (-17.197)
= 0.88V + 0.442V
= 1.322V

Therefore, the calculated Ecell for the electrochemical cell is 1.322V.

Answer 2

1.01 V

Step-by-step explanation:

From Nernst equation;

Ecell= E°cell- 0.0592/n log Q

Where;

Ecell= observed emf of the cell

E°cell= standard emf of the cell

n= number of moles of electrons transferred

Q= reaction quotient

Q= [Ag^+]^3/[MnO4^-] [H^+]^4

Q= [0.01]^3/[1.20] [1.50]^4

Q= 1.65×10^-7

Ecell= 0.88 - 0.0592/3 log 1.65×10^-7

Ecell= 0.88 - [0.0197×(-6.78)]

Ecell= 0.88 + 0.134

Ecell= 1.01 V