Questions in electrochemistry

SelectQuestion
For the redox reaction $Zn\,(s)+C{{u}^{2+}}(0.1\,M)\to Z{{n}^{2+}}(1M)+Cu\,(s)$ taking place in a cell, $E_{\text{cell}}^{0}$ is 1.10 volt. ${{E}_{\text{cell}}}$ for the cell will be $\left( 2.303\frac{RT}{F}=0.0591 \right)$
A solution containing one mole per litre of each $Cu{{(N{{O}_{3}})}_{2}},\,AgN{{O}_{3}},\,H{{g}_{2}}{{(N{{O}_{3}})}_{2}}$ and $Mg{{(N{{O}_{3}})}_{2}}$, is being electrolysed by using inert electrodes. The values of standard electrode potentials in volts (reduction potentials) are $Ag/A{{g}^{+}}=+0.80,$ $2Hg/Hg_{2}^{2+}=+0.79,$ $Cu/C{{u}^{2+}}=+0.34,\,Mg/M{{g}^{2+}}=-2.37$ with increasing voltage, the sequence of deposition of metals on the cathode will be
The standard reduction electrode potentials of four elements are $A = – 0.250 V$; $B = – 0.136 V$; $C = – 0.126 V$; $D = – 0.402 V$ The element that displaces $A$ from its compounds is
Calculate the electrode potential at $298\,K$ for $Zn|Z{{n}^{++}}$ electrode in which the activity of zinc ions is 0.001 M and $E_{Zn/Z{{n}^{++}}}^{0}$ is – 0.74 volts
$2{{H}^{+}}(aq)+2{{e}^{-}}\to {{H}_{2}}(g)$. The standard electrode potential for the above reaction is (in volts)
If the half cell reaction $A+{{e}^{-}}\to {{A}^{-}}$ has a large negative reduction potential, it follows that
For the electrochemical cell, $M|{{M}^{+}}||{{X}^{-}}|X,\,{{E}^{0}}({{M}^{+}}/M)=0.44\,V$ and ${{E}^{0}}(X/{{X}^{-}})=0.33\,V$. From this data one can deduce that
The standard cell potential of $Zn|Z{{n}^{2+}}_{(aq)}||C{{u}^{2+}}_{(aq)}|Cu$ cell is 1.10 V. The maximum work obtained by this cell will be
Electrode potentials of five elements A, B, C, D and E are respectively – 1.36, – 0.32, 0, – 1.26 and – 0.42. The reactivity order of these elements are in the order of
$Emf$ of a cell whose half cells are given below is $M{{g}^{2+}}+2{{e}^{-}}\to Mg\,(s)$; E = – 2.37 V $C{{u}^{2+}}+2{{e}^{-}}\to Cu\,(s)$; E = + 0.33 V

View Selected Questions (0)

Back to Categories

Back to Home