Materials Science And Engineering Properties
Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
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Chapter 10, Problem 10.6P

(a)

To determine

The reactions that occur at the two electrode.

(a)

Expert Solution
Check Mark

Explanation of Solution

Introduction:

The voltage developed in an electrochemical cell is directly related to the Gibbs free energy change of the reaction. The standard-electrode potentials are determined under equilibrium conditions at room temperature 25°C and 1 atmosphere of pressure. The reduction reactions occurs at the cathode and the oxidation reaction at the anode.

The copper reduction reaction is more electropositive than the iron reduction reaction.

Copper deposits on the copper electrode in accordance with the reduction reaction.

The copper reduction reaction or the reaction at cathode is given by,

  Cu2++2eCu

The oxidation reaction at the anode is given by,

  FeFe2++2e

Conclusion:

Thus, the reaction at the cathode is Cu2++2eCu and at anode is FeFe2++2e.

(b)

To determine

The voltage developed in the cell.

(b)

Expert Solution
Check Mark

Answer to Problem 10.6P

The voltage developed in the cell is 0.87V.

Explanation of Solution

Given:

The mass of copper in the solution is 635.4g.

The volume of solution is 1×103m3.

The mass of iron is 0.5585g.

Formula Used:

Write the expression for the concentration of copper.

  CCu=mCu63.54....... (I)

Here, CCu is the concentration of copper and mCu is mass of copper ion in the solution.

Write the expression for the concentration of iron.

  CFe=mFe55.85...... (II)

Here, CFe is the concentration of iron and mFe is mass of iron ion in the solution.

Write the expression for the half-cell voltage at the copper electrode.

  VCu=(V0)Cu+0.0592ZlogCCu...... (III)

Here, VCu is the half-cell voltage at the copper electrode and (V0)Cu is the voltage at copper electrode with 1-molarsolution.

Write the expression for the half-cell voltage at the iron electrode.

  VFe=(V0)Fe+0.0592ZlogCFe...... (IV)

Here, VFe is the half-cell voltage at the iron electrode and (V0)Fe is the voltage at iron electrode with 1-molarsolution.

Write the expression for the voltage developed in the cell.

  V=VCu+VFe...... (V)

Here, V is the voltage developed in the cell.

Calculation:

Substitute 635.4g for mCu in equation (I).

  CCu=635.463.54=10

Substitute 0.5585g for mFe in equation (II).

  CFe=0.558555.85=0.01

Substitute 0.340V for (V0)Cu , 2 for Z and 10 for CCu in equation (III).

  VCu=(0.340V)+0.05922log(10)=(0.340+0.0296)V=0.3696V

Substitute 0.44V for (V0)Fe , 2 for Z and 0.01 for CCu in equation (IV).

  VFe=(0.44V)+0.05922log(0.01)=(0.440.0592)V=0.4992V

Here, negative sign indicate the voltage at anode.

Substitute 0.3696V for VCu and 0.4992V for VFe in equation (V).

  V=0.3696V+0.4992V=0.8688V=0.87V

Conclusion:

Therefore, the voltage developed in the cell is 0.87V.

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ISBN:9781111988609
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