Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • Find the emf of the cell in which the following reaction takes place at <img alt="\mathrm{Ni}(\mathrm{s})+2 \mathr

Find the emf of the cell in which the following reaction takes place at 298 \mathrm{~K}
\mathrm{Ni}(\mathrm{s})+2 \mathrm{Ag}^{+}(0.001 \mathrm{M}) \rightarrow \mathrm{Ni}^{2}+(0.001 \mathrm{M})+2 \mathrm{Ag}(\mathrm{s})
(Given that  \mathrm{E^{0}_{cell} }=1.05\mathrm{V},\frac{2.303\: \mathrm{RT}}{\mathrm{F}}=0.0285\: \mathrm{at}\: 298\mathrm{K}  )
 

Option: 1

\mathrm{1.385\ V}


Option: 2

0.9615\: \mathrm{V}


Option: 3

1.05\: \mathrm{V}


Option: 4

1.0385\ \mathrm{V}


Answers (1)

best_answer

The cell reaction is

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

\mathrm{n=2 \text { and } Q=\frac{[\mathrm{Ni} ^{2+}]}{[\mathrm{Ag}^{+}]^{2}}}

\therefore From the nearest equation we have

\mathrm{E =E_{\text {cell }}^{0}-\frac{0.059}{2} \log \frac{\left[\mathrm{Ni}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}}} \\

    \mathrm{=1.05-0.0295 \log \frac{10^{-3}}{\left(10^{-3}\right)^{2}} }\\

    \mathrm{=1.05-0.0295 \log 10^{3}} \\

    =1.05-0.0295 \times 3 \\

    =0.9615 \mathrm{~V}

hence correct option is 2

Posted by

vishal kumar

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

'PM Modi personally supervising the situation': Centre tells SC in NEET case
anam-khan

NEET Paper Leak Row: NTA informs SC of security overhaul, institutional reforms
anam-khan

Re-NEET 2026 City Intimation Slip LIVE: NTA MBBS admit card soon; direct link, login details, exam date

Latest Question