What is the voltage of a galvanic cell made with silver and nickel?

1 Answer
Mar 11, 2016

Answer:

#E_(cell)^@=+1.05"V"#

Explanation:

List the standard electrode potentials from -ve to +ve:

#" "E^@("V")#

#stackrelcolor(blue)(leftarrow)color(white)(xxxxxxxxxxxxxx)#
#Ni_((aq))^(2+)+2erightleftharpoonsNi_((s))" "-0.25#

#Ag_((aq))^++erightleftharpoonsAg_((s))" "+0.8#

#stackrelcolor(red)(rightarrow)color(white)(xxxxxxxxxxxxxx)#
You can see that the #"Ag"^+"/""Ag"# 1/2 cell has the most +ve #E^@# value so this will take in electrons and move left to right.

The #"Ni"^(2+)"/""Ni"# 1/2 cell will, therefore, take in these electrons and move right to left.

The voltage of the cell is an empirically measured quantity so must always have a +ve value. So to calculate #E_(cell)^@# subtract the least +ve #E^@# from the most +ve #E^@# value#rArr#

#E_(cell)^@=+0.8-(-0.25)=+1.05"V"#