Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The equilibrium constant for the reaction <img alt="\mathrm{Zn}(\mathrm{s})+\mathrm{Sn}^{2+}(\mathrm{aq}) \rightleftharpoons \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Sn}(\mathrm{s})$ is $1 \

The equilibrium constant for the reaction

\mathrm{Zn}(\mathrm{s})+\mathrm{Sn}^{2+}(\mathrm{aq}) \rightleftharpoons \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Sn}(\mathrm{s})$ is $1 \times 10^{20} at 298 \mathrm{~K}. The magnitude of standard electrode potential of \mathrm{Sn} / \mathrm{Sn}^{2+}$ if $\mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V} is ______________ \times 10^{-2} \mathrm{~V} (Nearest integer).
Given : \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}

Option: 1

17


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

Given

\begin{aligned} & \mathrm{Zn}(\mathrm{s})+\mathrm{Sn}^{2+}(\mathrm{aq} .) \rightleftharpoons \mathrm{Zn}^{2+}(\text { aq. })+\mathrm{Sn}(\mathrm{s}) \\ & \mathrm{K}_{\mathrm{C}}=1 \times 10^{20} \\ & \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V} \\ & \mathrm{E}_{\mathrm{cell}}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.059}{\mathrm{n}} \log _{10} \mathrm{~K}_{\mathrm{c}} \end{aligned}

\begin{aligned} & 0=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.059}{2} \times 20 \\ & \mathrm{E}_{\text {cell }}^{\circ}=0.59 \\ & \mathrm{E}_{\text {cell }}^{\circ}=\underset{(\mathrm{RP})}{\mathrm{E}_{\text {Cathode }}^{\circ}}-\underset{(\mathrm{RP})}{\mathrm{E}_{\text {Anode }}^{\circ}} \\ & \end{aligned}

\begin{aligned} & 0.59-\mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\circ}-\mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ} \\ & 0.59=\mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\circ}-(-0.76) \\ & \mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\circ}=0.17 \\ & \mathrm{E}_{\mathrm{Sn} / \mathrm{Sn}^{2+}}^{\circ}=17 \times 10^{-2} \end{aligned}

Posted by

Kshitij

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Trending Articles/News

anam-khan

AESL expands Aakash Digital 2.0 to offer AI-powered coaching for NEET, JEE, Olympiads
anam-khan

Bihar Board extends BSEB Super 50 free coaching application last date to March 26
anam-khan

JEE Main City Intimation Slip 2025 Live: NTA session 2 exam city slips soon at jeemain.nta.nic.in

Latest Question