1. Tardigrade
  2. Question
  3. Chemistry
  4. Electrode potential for Ag electrode varies according to the equation EA g+ / A g=EA g+ / A g°-(0.059/1) log (1/[A g+]) If the graph of E Ag + / Ag versus log [ Ag +]is plotted, the value of slope is

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Q. Electrode potential for $Ag$ electrode varies according to the equation
$E_{A g^{+} / A g}=E_{A g^{+} / A g}^{\circ}-\frac{0.059}{1} \log \frac{1}{\left[A g^{+}\right]}$
If the graph of $E _{ Ag ^{+} / Ag }$ versus $\log \left[ Ag ^{+}\right]$is plotted, the value of slope is_______

Electrochemistry

Solution:

$E _{ Ag ^{+} / Ag }= E _{ Ag ^{+} / Ag }^{\circ}-\frac{0.059}{1} \log \frac{1}{\left[ Ag ^{+}\right]}$
$\therefore E _{ Ag ^{+} / Ag }= E _{ Ag ^{+} / Ag }^{0}+\frac{0.059}{1} \log \left[ Ag ^{+}\right]$
This is the general form of $y=m x+c$.
$\therefore $ Slope $= m =0.059$