An electrochemical cell consists of a silver metal electrode immersed in a solution with [Ag + ] = 1.00 M separated by a porous disk from a compartment with a copper metal electrode immersed in a solution of 10.00 M NH 3 that also contains 2.4 × 10 −3 M Cu(NH 3 ) 4 2+ . The equilibrium between Cu 2+ and NH 3 is: Cu 2 + ( a q ) + 4 NH 3 ( a q ) ⇌ Cu ( N H 3 ) 4 2 + ( a q ) K = 1.0 × 10 13 and the two cell half-reactions are: Assuming Ag + is reduced, what is the cell potential at 25°C?

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Interpretation Introduction Interpretation: An electrochemical cell consisting of Silver and Copper electrode dipped in Ammonia solution and its equilibrium constant is given. The value of cell potential at 25 ° C for the given cell is to be calculated. Concept introduction: The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation. The value of E cell is calculated using Nernst formula, E = E ° − ( R T n F ) ln ( Q ) At room temperature, the above equation specifies as, E = E ° − ( 0.0591 n ) log ( Q ) To determine: The value of cell potential at 25 ° C for the given cell. Explanation Given, The concentration of Ag + is 1.00 M . The concentration of NH 3 solution is 1.00 M . The concentration of NH 3 solution is 1.00 M . The concentration of [ Cu ( NH 3 ) 4 ] 2 + solution is 2.4 × 10 − 3 M . The value of the equilibrium constant is 1.0 × 10 13 The equilibrium reaction is, Cu 2 + ( a q ) + 4 NH 3 ( a q ) → [ Cu ( NH 3 ) 4 ] 2 + ( a q ) The reaction taking place at the cathode is, Ag + + e − → Ag E ° red = 0.80 V The reaction taking place at the anode is, Cu → Cu 2 + + 2 e − E ° OX = − 0.34 V Multiply reduction half-reaction with a coefficient of 2 and then add both the reactions as, 2Ag + + 2 e − → 2 Ag Cu → Cu 2 + + 2 e − The final equation is, 2Ag + + Cu → 2 Ag+Cu 2 + The value of E ° cell is given as, E ° cell = E ° ox + E ° red Where, E ° ox is the oxidation potential of the electrode. E ° red is the reduction potential of the electrode. Substitute the values of E ° ox and E ° red in the above equation as, E ° cell = E ° ox + E ° red = − 0

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Chemistry: An Atoms First Approach

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Chemistry: An Atoms First Approach

2nd Edition

Steven S. Zumdahl + 1 other

Publisher: Cengage Learning

ISBN: 9781305079243

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Interpretation Introduction

Interpretation:

An electrochemical cell consisting of Silver and Copper electrode dipped in Ammonia solution and its equilibrium constant is given. The value of cell potential at 25 °C for the given cell is to be calculated.

Concept introduction:

The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation.

The value of Ecell is calculated using Nernst formula,

E=E°−(RTnF)ln(Q)

At room temperature, the above equation specifies as,

E=E°−(0.0591n)log(Q)

To determine: The value of cell potential at 25 °C for the given cell.

Explanation of Solution

Given,

The concentration of Ag+ is 1.00 M .

The concentration of NH3 solution is 1.00 M .

The concentration of [Cu(NH3)4]2+ solution is 2.4×10−3 M .

The value of the equilibrium constant is 1.0×1013

The equilibrium reaction is,

Cu2+(aq)+4NH3(aq)→[Cu(NH3)4]2+(aq)

The reaction taking place at the cathode is,

Ag++e−→Ag E°red=0.80 V

The reaction taking place at the anode is,

Cu→Cu2++2e− E°OX=−0.34 V

Multiply reduction half-reaction with a coefficient of 2 and then add both the reactions as,

2Ag++2e−→2Ag Cu→Cu2++2e−

The final equation is,

2Ag++Cu→2Ag+Cu2+

The value of E°cell is given as,

E°cell=E°ox+E°red

Where,

E°ox is the oxidation potential of the electrode.

E°red is the reduction potential of the electrode.

Substitute the values of E°ox and E°red in the above equation as,

E°cell=E°ox+E°red= −0

Chemistry: An Atoms First Approach

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Chemistry: An Atoms First Approach

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