# E.M.F of cell Ni|Ni+2(0.1)||Au+3(1.0)|Au is ...,if E° for Ni+2|Ni is -0.25V,E° for Au+3|Au is 1.50V ?

Jun 3, 2018

$E \left(c e l l\right) = 1.75 \textcolor{w h i t e}{l} \text{V}$

#### Explanation:

This question defines the voltaic cell with a cell notation. This notation can tell plenty of information about the structure and the reactions going on within the cell.

The standard cell notation indicates the respective composition of and reaction going on at each of the electrodes, respectively. By convention, chemical compositions of the anode- the electrode where oxidation takes place- are written on the left-hand side of the double vertical line that symbolizes a salt bridge, whereas, compositions of the cathode are on the right-hand side of the double lines. That is:

$\textcolor{red}{\text{ Anode")||color(darkblue)("Cathode}}$

For this particular setup,

$\textcolor{red}{\text{Ni") | color(red)("Ni"^(2+))(aq,0.1color(white)(l)"M") || color(darkblue)("Au"^(+3))(aq, 1.0color(white)(l)"M")|color(darkblue)("Au}}$

Thus

E(color(red)("anode"))=E(color(red)("Ni") | color(red)("Ni"^(2+))(aq,0.1color(white)(l)"M"))=-0.25color(white)(l)"V"
E(color(darkblue)("cathode"))=E(color(darkblue)("Au"^(+3))(aq, 1.0color(white)(l)"M")|color(darkblue)("Au"))=1.50color(white)(l)"V"

The cell potential of voltaic cells is equal to the difference between the electrode potential of the cathode and the anode.

E("cell")=E(color(darkblue)("cathode"))-E(color(red)("anode"))
$\textcolor{w h i t e}{E \left(\text{cell"))=color(darkblue)(1.50color(white)(l)"V")-(color(red)(-0.25color(white)(l)"V}\right)}$
color(white)(E("cell")) =1.75color(white)(l)"V"

Reference
"17.8: Cell Notation and Conventions", Chemistry Libretext, https://chem.libretexts.org/Textbook_Maps/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/17Electrochemical_Cells/17.08%3A_Cell_Notation_and_Conventions