   Chapter 18, Problem 113AE

Chapter
Section
Textbook Problem

# Gold is produced electrochemically from an aqueous solution of Au ( CN ) 2 − containing an excess of CN–. Gold metal and oxygen gas are produced at the electrodes. What amount (moles) of O2 will be produced during the production of 1.00 mole of gold?

Interpretation Introduction

Interpretation:

The amount (moles) of oxygen produced during the production of 1.00 mole of gold is to be calculated.

Concept introduction:

The number of moles is defined as the ratio of mass with the molecular mass of an element. The mass of an element is the amount of the substance present in an element. The oxidation reaction is defined as the reaction in which loss of electrons takes place and in reduction the gain of electrons occurs.

To determine: The amount (moles) of oxygen produced during the production of 1.00 mole of gold.

Explanation

Given

The moles of gold is 1.00 mole .

The reduction and oxidation reaction takes place in the cell are,

Reduction: Au(CN)2+eAu+2CN (1)

Oxidation: 2H2OO2+4H++4e (2)

The overall reaction by multiplying the reduction reaction by 4 and adding oxidation reaction on cancelling electrons is,

4Au(CN)2+2H2O4Au+8CN+O2+4H+ (3)

In the above reaction number of moles of gold consumed is four times than the number of moles of oxygen

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