1. Tardigrade
  2. Question
  3. Chemistry
  4. Consider the cell underset1 atmH2 (Pt)|. undersetpH = 5 . 00H3 O+ (aq)|.|. undersetx M(Ag)+|.Ag The measured EMF of the cell is 1.0V . What is the value of x? EAg+ , Ag0=+0.8V [T = 25 ° C] [(2 . 303 RT/F) = 0 . 06 , log 0 . 02 = - (5/3)] Give your answer after multiplying 100 and round off to the nearest integer.

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Q. Consider the cell $\underset{1 atm}{H_{2} \left(Pt\right)}\left|\right.\underset{pH = 5 . 00}{H_{3} O^{+} \left(aq\right)}\left|\right.\left|\right.\underset{x M}{\left(Ag\right)^{+}}\left|\right.Ag$
The measured $EMF$ of the cell is $1.0V$ . What is the value of $x?$ $E_{Ag^{+} , Ag}^{0}=+0.8V$ $\left[T = 25 ^\circ C\right]$
$\left[\frac{2 . 303 RT}{F} = 0 . 06 , log 0 . 02 = - \frac{5}{3}\right]$
Give your answer after multiplying 100 and round off to the nearest integer.

NTA AbhyasNTA Abhyas 2022

Solution:

Reaction for the cell is
$\frac{1}{2}H_{2}+Ag^{+} \rightarrow H^{+}+Ag$
$E_{cell}=E_{cell}^{o}-\frac{0 . 06}{n}logQ1.0=0.80-\frac{0 . 06}{1}log\frac{10^{- 5}}{x}log\frac{10^{- 5}}{x}=-\frac{0 . 2}{0 . 06}-5-logx=-\frac{10}{3}logx=-5+\frac{10}{3}=-\frac{5}{3}x=0.02$