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Question

Mr.Nernst had an apprentice named Electra. She is hard-working but is not very good with Chemistry. She needs a lot of help from you to pass as an apprentice. Please help the damsel in distress!

One day Mr.Nernst told her to find the EMF of the following cell using experiments at 25∘C

Mg(s)+2Ag+(0.0001M)→Mg2+(0.130M)+2Ag(s)

Electra reported the value to be +2.6V. Mr.Nernst told her to recheck the value using his theory, knowing Standard EMF = 3.17 V.

What was the right value? Round off the answer to nearest integer.

Type your answer here.___

One day Mr.Nernst told her to find the EMF of the following cell using experiments at 25∘C

Mg(s)+2Ag+(0.0001M)→Mg2+(0.130M)+2Ag(s)

Electra reported the value to be +2.6V. Mr.Nernst told her to recheck the value using his theory, knowing Standard EMF = 3.17 V.

What was the right value? Round off the answer to nearest integer.

Type your answer here.

Open in App

Solution

Let's help Electra together.

First, write down the Nernst equation's general form for this reaction. Observe that I haven't included the concentrations of Mg(s) and Ag(s) since these are taken as unity.

Ecell=E∘−2.303RTnFlog([Mg2+][Ag2+]2)

E∘=3.17 V as given in the question.

R=8.314 JK−1mol−1

T=25∘C=298 K

Also, [Mg2+]=0.310 M

[Ag+]=0.0001 M or 10−4 M

To find n, we need to write down the half-reactions of the cell and balance the electrons. Let's do that. Let's take the equation given as reference. We can see that,

Mg→Mg2++2e−

2Ag++2e−→2Ag

So, we see that the reactions is balanced and n = 2

By plugging in all these values, we get

Ecell=E∘−2.303RTnFlog([Mg2+][Ag2+]2)

Ecell=E0−2.303∗8.314∗298n∗96500log(0.1310−3)

If we calculate the value, we'll get Ecell=2.96V. We can round that off to 3 V

First, write down the Nernst equation's general form for this reaction. Observe that I haven't included the concentrations of Mg(s) and Ag(s) since these are taken as unity.

Ecell=E∘−2.303RTnFlog([Mg2+][Ag2+]2)

E∘=3.17 V as given in the question.

R=8.314 JK−1mol−1

T=25∘C=298 K

Also, [Mg2+]=0.310 M

[Ag+]=0.0001 M or 10−4 M

To find n, we need to write down the half-reactions of the cell and balance the electrons. Let's do that. Let's take the equation given as reference. We can see that,

Mg→Mg2++2e−

2Ag++2e−→2Ag

So, we see that the reactions is balanced and n = 2

By plugging in all these values, we get

Ecell=E∘−2.303RTnFlog([Mg2+][Ag2+]2)

Ecell=E0−2.303∗8.314∗298n∗96500log(0.1310−3)

If we calculate the value, we'll get Ecell=2.96V. We can round that off to 3 V

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