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Question: A button cell used in watches functions as following: \(Zn(s)+A{{g}_{2}}O(s)+{{H}_{2}}O\Leftrighta...

A button cell used in watches functions as following:
Zn(s)+Ag2O(s)+H2O⇔2Ag(s)+Zn2+(aq)+2OH−(aq)Zn(s)+A{{g}_{2}}O(s)+{{H}_{2}}O\Leftrightarrow 2Ag(s)+Z{{n}^{2+}}(aq)+2O{{H}^{-}}(aq)
If half cell potentials are:
Zn2+(aq)+2e−→Zn(s),E0=−0.76V Ag2O(s)+H2O(l)+2e−→2Ag(s)+2OH−(aq),E0=0.34V \begin{aligned} & Z{{n}^{2+}}(aq)+2{{e}^{-}}\to Zn(s),{{E}^{0}}=-0.76V \\\ & A{{g}_{2}}O(s)+{{H}_{2}}O(l)+2{{e}^{-}}\to 2Ag(s)+2O{{H}^{-}}(aq),{{E}^{0}}=0.34V \\\ \end{aligned}
The cell potential will be:
A. 1.10 V
B. 0.42 V
C. 0.84 V
D. 1.34 V

Explanation

Solution

To solve the given problem at first we should know about some basic formulas:
Nernst equation:
E=E0−(2.303 RT÷nF)log kcE={{E}_{0}}-\left( 2.303\text{ }RT\div nF \right)log\text{ }{{k}_{c}}
or $$$$$$E={{E}{0}}-(0.059\div n)log\text{ }{{k}{c}} where, $E$= cell potential {{E}{0}}=standard electrode potential $R$= universal gas constant=8.3144598\text{ }J.\text{ }mo{{l}^{-1}}.\text{ }{{K}^{-1}} $T$= temperature $n$=number of moles of electrons transferred in the balanced equation $F$= Faraday constant=96,485\text{ }Cmo{{l}^{-1}} {{K}{c}}=equilibriumconstant= equilibrium constant {{E}{0}}=reductionpotentialofcathode−reductionpotentialofanodeor,= reduction potential of cathode -reduction potential of anode or, {{E}{0}}$$=oxidation potential of anode + reduction potential of cathode

Complete step by step solution:
Standard electrode potential: The potential difference developed between metal electrode and the solution of its ions of unit molarity at 298 K is called standard electrode potential.
In galvanic cells cathode is positively charged and anode is negatively charged . Current flows from cathode to anode.
Reduction takes place at cathode and oxidation takes place at anode.
Now let's come to the actual solution,
We will be using the nernst equation to solve the problem,
First we need to find the E0{{E}_{0}} value from the formula,
E0{{E}_{0}}= R.PR.P ofof cathodecathode −- R.PR.P ofof anodeanode
From the half cell potentials we will get the reduction potential values
(O.34−(−0.76))V  =1.1V\left( O.34-\left( -0.76 \right) \right)V~~=1.1V
=(O.34−(−0.76))V=(O.34-(-0.76))V
=1.1V=1.1V
Now here E=E= E0{{E}_{0}}

So the cell potential is 1.1V which matches the first option that is option A.

Note: Electrochemistry deals with the study of the chemical changes which occur on passing electric current into certain chemical systems and also with the generation of electricity by carrying chemical reactions. Electrochemical cell is a cell in which electrical energy is generated as a result of redox reaction carried in the cell. Electrons flow from anode to cathode along the external circuit while current flows in the opposite direction in an electrochemical cell. Ions flow from one half cell to the other via salt bridge.