   Chapter 18, Problem 46E

Chapter
Section
Textbook Problem

# Give the balanced cell equation and determine for the galvanic cells based on the following half-reactions. Standard reduction potentials are found in Table 17-1.a. Cr 2 O 7 2 − + 14H + + 6e − → 2Cr 3 + + 7H 2 O H 2 O 2 + 2 H + + 2e − → 2H 2 O b. 2H + +2e − → H 2 Al 3+ +3e − → Al

(a)

Interpretation Introduction

Interpretation:

The equations for two galvanic cells are given. The balanced cell equation and the value of E° for each cell is to be stated.

Concept introduction:

The species at the anode undergoes oxidation while the species at the cathode undergoes reduction. The species with higher reduction potential undergoes reduction while the species with lower reduction potential undergoes oxidation.

To determine: The balanced cell equation and E° for the given galvanic cell.

Explanation

The balanced cell equation is,

3H2O2+2Cr3++H2OCr2O72+8H+

The value of E°cell is 0.45V_ .

The reaction taking place at cathode is,

H2O2+2H++2e2H2OE°red=+1.78V

The reaction taking place at anode is,

2Cr3++7H2OCr2O72+14H++6eE°ox=1

(b)

Interpretation Introduction

Interpretation:

The equations for two galvanic cells are given. The balanced cell equation and the value of E° for each cell is to be stated.

Concept introduction:

The species at the anode undergoes oxidation while the species at the cathode undergoes reduction. The species with higher reduction potential undergoes reduction while the species with lower reduction potential undergoes oxidation.

To determine: The balanced cell equation and E° for the given galvanic cell.

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