Question

Given,
$\begin{align}
  & {{E}^{\circ }}(N{{i}^{2+}}/Ni)=-0.25V \\
 & {{E}^{\circ }}(A{{u}^{3+}}/Au)=1.50V \\
\end{align}$
 The EMF of the voltaic cell.
$Ni/N{{i}^{2+}}(1.0M)//A{{u}^{3+}}(1.0M)/Au$
A.1.25V
B.-1.75V
C.1.75V
D.4.0V

Answer 1

Hint: Before solving this, we should know that The maximum potential difference between two electrodes of a cell is known as EMF of a cell or Electromotive force of a cell. The expression for it is ${{E}_{cell}}={{E}^{\circ }}_{cell}-\dfrac{RT}{nF}\ln Q$. Now, we can apply this formula and put all the values in it.

Complete answer:
The net voltage between the oxidation and reduction half-reactions is also the EMF of a cell. It is also used to see if the electrochemical cell is galvanic or not.
$aA+bB\xrightarrow{ne}cC+dD$
${{E}_{cell}}={{E}^{\circ }}_{cell}-\dfrac{RT}{nF}\ln Q$
So, Nernst gave us an equation :
$EMF={{E}^{\circ }}_{cell}-\dfrac{0.059}{6}\log \dfrac{(N{{i}^{2+}})}{(A{{u}^{2+}})}$
$Ni/N{{i}^{2+}}(1.0M)//A{{u}^{3+}}(1.0M)/Au$
At cathode, there will be reduction-
$[A{{u}^{3+}}+3{{e}^{-}}\to Au(s)]\times 2$
At anode, there will be oxidation-
$[Ni(s)\to N{{i}^{2+}}+2{{e}^{-}}]\times 3$
$3Ni(s)+2A{{u}^{3+}}\to 2Au(s)+3N{{i}^{2+}}$

${{E}^{\circ }}_{cell}={{E}_{cathode}}-{{E}_{anode}}$
            =1.5+0.25
            =1.75 V
$EMF={{E}^{\circ }}_{cell}-\dfrac{0.059}{6}\log \dfrac{(N{{i}^{2+}})}{(A{{u}^{2+}})}$
            $=1.75-\dfrac{0.059}{6}\log \dfrac{{{1}^{3}}}{{{1}^{2}}}$
             $=1.75-\dfrac{0.059}{6}\log 1$
                         (log 1 = 0)
 EMF = 1.75 V

So, Option (C) 1.75V is correct.

Note:
ELECTROCHEMICAL CELL
The device used for the generation of electricity from a chemical reaction is an electrochemical cell. The conversion of chemical energy into electrical energy by the device. There is an exchange of electrons that happens in the chemical reaction which is necessary for an electrochemical cell to operate. These reactions are known as redox reactions. A cell is distinguished by its voltage. The generation of the same voltage happens irrespective of the size of the cell. The chemical composition is dependent on the cell voltage in which the cell operates at ideal conditions. Some factors like temperature difference, change in concentration plays a role in the difference in cell voltage from its ideal value. Nernst Equation is the equation that is used to obtain the EMF value of a cell in which the standard cell potential of a cell is given. Walther Nernst formulated the Nernst equation.