   Chapter 18, Problem 39E

Chapter
Section
Textbook Problem

# Calculate values for the galvanic cells in Exercise 37.

(a)

Interpretation Introduction

Interpretation:

Two galvanic cells with two different chemical reactions are given. The value of E° for both the cells is to be calculated.

Concept introduction:

The galvanic cell converts chemical energy into electrical energy while the electrolytic cell converts electrical energy into chemical energy.

The species at the anode undergoes oxidation while the species at cathode undergoes reduction and the electrons generated at the anode are transferred through wire to the cathode.

Explanation

The value of E°cell is calculated as 0.03V_ .

The reaction taking place at cathode is,

Cl2(g)+2e-2Cl-(aq)red=+1.36V

The reaction taking place at anode is,

2Cr3+(aq)+7H2O(l)Cr2O72-(aq)+14H+(aq)+6e-ox=-1.33V

Multiply reduction half reaction with the coefficient of 3 and both the reduction and oxidation half-reaction,

3Cl2(g)+6e-6Cl-(aq)2Cr3+

(b)

Interpretation Introduction

Interpretation:

Two galvanic cells with two different chemical reactions are given. The value of E° for both the cells is to be calculated.

Concept introduction:

The galvanic cell converts chemical energy into electrical energy while the electrolytic cell converts electrical energy into chemical energy.

The species at the anode undergoes oxidation while the species at cathode undergoes reduction and the electrons generated at the anode are transferred through wire to the cathode.

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