# Calculate the pH of the cathode compartment for the following reaction given = 3.01 V when [Cr 3+ ] = 0.15 M , [Al 3+ ] = 0.30 M, and [Cr 2 O 7 2− ] = 0.55 M . 2Al ( s ) + Cr 2 O 7 2 − ( a q ) + 14H + ( a q ) → 2A1 3 + ( a q ) + 2Cr 3 + ( a q ) + 7H 2 O ( l )

### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

Chapter
Section

### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 17, Problem 72E
Textbook Problem
1 views

## Calculate the pH of the cathode compartment for the following reaction given  = 3.01 V when [Cr3+] = 0.15 M, [Al3+] = 0.30 M, and [Cr2O72−] = 0.55 M. 2Al ( s )   +  Cr 2 O 7 2 − ( a q ) +  14H + ( a q ) →  2A1 3 + ( a q ) +  2Cr 3 + ( a q ) +  7H 2 O ( l )

Interpretation Introduction

Interpretation:

The chemical reaction for a galvanic cell is given. The pH of the cathode compartment for the given reaction is to be calculated.

Concept introduction:

The H+ ion concentration is measured by the value of pH . Higher pH corresponds to a basic solution while the lower value corresponds to acidic solution. The value of pH is also demonstrated using the Nernst equation.

The Nernst formula is defined as follows,

E=E°(RTnF)ln(Q)

To determine: The value of E at 25°C for this battery at the given concentrations of different ions.

### Explanation of Solution

Given

The value of concentration of Cr3+ is 0.15M .

The value of concentration of Al3+ is 0.30M .

The value of concentration of Cr2O72 is 0.55M .

The value of Ecell is 3.01V .

The reaction taking place at cathode is,

Cr2O72(aq)+14H+(aq)+6e2Cr3+(aq)+7H2O(l)E°red=1.33V

The reaction taking place at anode is,

AlAl3++3eE°ox=1.66V

Multiply the oxidation half with a coefficient of 2 and then add both the oxidation and reduction half reaction,

Cr2O72(aq)+14H+(aq)+6e2Cr3+(aq)+7H2O(l)2Al2Al3++6e

The final equation is,

Cr2O72(aq)+2Al(s)+14H+(aq)2Cr3+(aq)+2Al3+(aq)+7H2O(l)

The value of E°cell is calculated as,

E°cell=E°ox+E°red=1.66V+1.33V=2.99V_

The reaction involves the transfer of 6 moles of electrons and the value of Ecell=3

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