1) Write a balanced chemical equation for the overall reaction represented by th

Question

1)

Write a balanced chemical equation for the overall reaction represented by the cell notation below.

Cr(s) | Cr3+(aq) || Br–(aq) | Br2(g) | Pt(s)

2Cr3+(aq) + 6Br–(aq) ? 2Cr(s) + 3Br2(g)

Cr(s) + Cr3+(aq) ? Br–(aq) + Br2(g)

Cr(s) + 3Br2(g) ? Cr3+(s) + 2Br–(aq)

Cr(s) + 2Br–(aq) ? Br2(g) + Cr3+(aq)

2Cr(s) + 3Br2(g) ? 2Cr3+(aq) + 6Br–(aq)

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2)

Consider the following half-reactions:

Which of the above metals or metal ions will oxidize Pb(s)?

Cu2+(aq) and Fe2+(aq)

Ag(s) and Cu(s)

Ag+(aq) and Cu2+(aq)

Fe(s) and Al(s)

Fe2+(aq) and Al3+(aq)

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3)

What is the balanced spontaneous reaction and standard cell potential of an electrochemical cell constructed from half cells with the following half reactions?

Pt2+(aq) + 2e? ? Pt(s) E° = 1.180 V

Pb2+(aq) + 2e? ? Pb(s) E° = –0.130 V

2Cr3+(aq) + 6Br–(aq) ? 2Cr(s) + 3Br2(g)

Cr(s) + Cr3+(aq) ? Br–(aq) + Br2(g)

Cr(s) + 3Br2(g) ? Cr3+(s) + 2Br–(aq)

Cr(s) + 2Br–(aq) ? Br2(g) + Cr3+(aq)

2Cr(s) + 3Br2(g) ? 2Cr3+(aq) + 6Br–(aq)

Explanation / Answer

Cr(s) | Cr3+(aq) || Br–(aq) | Br2(g) | Pt(s)

Cr(s) -----> Cr3+(aq) + 3e- .......1

2 Br–(aq) + 2e- ------> Br2(g) .........2

To balance out the number of electrons multiply eqn 1 by 2 and equation 2 by 3.

Thus we will get,

2Cr(s) -----> 2Cr3+(aq) + 6e- .......1'

6Br–(aq) + 6e- ------> 3Br2(g) .........2'

After adding 1' + 2' , we will get 2Cr3+(aq) + 6Br–(aq) 2Cr(s) + 3Br2(g) as a balanced reaction.

2)

The whoever metal have more reduction potential than that of Pb will feasibly oxidises the Pb.

Cu2+(aq) + 2 e– Cu(s)

Therefore only these two metal can oxidises Pb.

3.

Pt2+(aq) + 2e Pt(s) E° = 1.180 V ------1

Pb2+(aq) + 2e Pb(s) E° = –0.130 V -----2

reverse the equation second as it have negative potential and it was then added to equation 1,

we will get,

Pt2+(aq) + Pb(s) Pt(s) + Pb2+(aq) = 1.310 V as a balanced equation.

E° = +0.34 V ,

Pb2+(aq) + 2 e– Pb(s) E° = –0.13 V

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