Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A galvanic cell is constructed by combining two half-cells. One half-cell consists of a nickel metal electrode immersed in a <img alt="0.10 \mathrm{M} \; \; \mathrm{NiSO}_4" src="https://

A galvanic cell is constructed by combining two half-cells. One half-cell consists of a nickel metal electrode immersed in a 0.10 \mathrm{M} \; \; \mathrm{NiSO}_4 solution, while the other half-cell consists of a silver metal electrode immersed in a 1.0 \mathrm{M}\; \mathrm{AgNO}_3 solution. At 298 K, the measured cell potential is 0.32 V. What is the standard cell potential for this reaction?

Option: 1

\begin{aligned} & 0.59 \mathrm{~V} \\ \end{aligned}


Option: 2

\begin{aligned} & 0.26 \mathrm{~V} \\ \end{aligned}


Option: 3

\begin{aligned} & -0.32 \mathrm{~V} \\ \end{aligned}


Option: 4

\begin{aligned} & -0.59 \mathrm{~V} \end{aligned}


Answers (1)

best_answer

The given galvanic cell can be represented as:

\mathrm{Ni}(\mathrm{s})\left|\mathrm{Ni}^{2+}(\mathrm{aq}, 0.10 \mathrm{M})|| \mathrm{Ag}^{+}(\mathrm{aq}, 1.0 \mathrm{M})\right| \mathrm{Ag}(\mathrm{s})

The measured cell potential (E) is 0.32 V. The standard cell potential \mathrm{E^{\circ}} can be determined using the Nernst equation:

E=E^{\circ}-\frac{0.0592 \mathrm{~V}}{n} \log Q

where n is the number of moles of electrons transferred in the balanced chemical equation and Q is the reaction quotient.

For the given cell, the balanced chemical equation is:

\mathrm{Ni}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})

The number of moles of electrons transferred in this equation is 2. Therefore, n = 2.

The reaction quotient Q can be expressed as:

Q=\frac{\left[\mathrm{Ni}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^2}

Substituting the given concentrations, we get:

Q=\frac{0.10 \mathrm{M}}{(1.0 \mathrm{M})^2}=0.10

Substituting the values of E, n, and Q in the Nernst equation, we get:

0.32 \mathrm{~V}=E^{\circ}-\frac{0.0592 \mathrm{~V}}{2} \log (0.10)

Solving for E^{\circ}, we get:

E^{\circ}=0.26 \mathrm{~V}

Therefore, the correct option is 2.

Posted by

vinayak

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question