   Chapter 19.2, Problem 19.3CYU

Chapter
Section
Textbook Problem

Describe how to set up a voltaic cell using the following half-reactions:Reduction half-reaction:    Ag+(aq) + e− → Ag(s)Oxidation half-reaction:    Ni(s) → Ni2+(aq) + 2 e−Which electrode is the anode, and which is the cathode? What is the overall cell reaction? What is the direction of electron flow in an external wire connecting the two electrodes? Describe the ion flow in a salt bridge (with NaNO3) connecting the cell compartments.

Interpretation Introduction

Interpretation:

The anode, cathode, overall reaction and the direction of electron flow in the external wire has to be determined.

Reduction: Ag+(aq)+e- Ag(s)Oxidation: Ni(s) Ni2+(aq)+2e-

Concept introduction:

Voltaic cell or Galvanic cell:

The device to produce electricity by using chemical reactions. In these divces are redox chemical reactions are occured.

A voltaic cell converts chemical energy into electrical energy.

It consists of two half cells. Each half cell consists of a metal and a solution of a salt of metal. Two half cells are connected by salt bridge.

The chemical reaction in the half cell is an oxidation reduction (redox)reactions.

For example:

Cell diagram of voltaic or galvanic cell is as follows.

Salt bridge                        Cu(s)|Cu2+(aq)  ||  Ag+(aq)|Ag(s)____________     ___________                                       Half cell             Half cell

Explanation

The given half cell reaction is as follows.

Reduction: Ag+(aq)+e- Ag(s)Oxidation: Ni(s) Ni2+(aq)+2e-

In the voltaic cell has two half cells. In this cell oxidation occurs at anode and reduction occurs at cathode.

From the given reaction, Ni atom increases their oxidation state 0 to +2.Therefore, it is an oxidation reaction and occurs at anode.

Ag atom reduces their oxidation state +1 to 0. Therefore, it is a reduction reaction occurs at cathode.

The Ni electrode is an anode and Ag electrode is cathode.

And the electrons move through salt bridge From Fe electrode to Cu electrode

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