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What is the potential for the cell Cr|...

What is the potential for the cell
`Cr|Cr^(3+)(0.1M)||Fe^(2+)(0.01M)|Fe`
`E^(@)Cr^(3+)// Cr=-0.74V`,
`E^(@)Fe^(2+)//Fe=-0.44V`

Text Solution

AI Generated Solution

To calculate the potential for the cell `Cr|Cr^(3+)(0.1M)||Fe^(2+)(0.01M)|Fe`, we will follow these steps: ### Step 1: Identify the Anode and Cathode - In the given cell notation, the left side represents the anode and the right side represents the cathode. - Anode: `Cr|Cr^(3+)` - Cathode: `Fe^(2+)|Fe` ### Step 2: Write the Half-Reactions ...
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Calculate the e.f.m of the cell Cr|Cr^(3+)(0.1M)||Fe^(2+)(0.01M)|Fe [given that E_(Cr^(3+)//Cr)^(@)=-0.75,E_(Fe^(2+)//Fe)^(@)=-0.45V]

Calculate the emf of the cell Cr∣ Cr^(3+)(0.1 M)∣∣Fe^(2+) (0.01M)∣Fe (Given: E^(@)Cr^(3+) /Cr =−0.75 volt; E^(@)Fe^(2+) /Fe =−0.45 volt).

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For the following cell, calculate the emf: Al//Al^(3+) (0.01 M)"||" Fe^(2+) (0.02M)"|"Fe Given: E_(Al^(3+)//Al)^(Theta) = -1.66V, E_(Fe^(2+)//Fe)^(Theta)= -0.44V

Calculate the emf of the cell Fe(s)+2H^(+)(1M)toFe^(+2)(0.001M)+H_(2)//(g),1 atm) (Given : E_(Fe^(+2)//Fe)=-0.44V)

The emf of the cell Cr(s) abs(Cr^(3+)(1.0M))abs(Co^(2+)(1.0M))Co(s) [E^(o)" For " Cr^(3+)abs(Cr(s)=-0.74V & Co^(2+))Co(s) -0.28V]:

The standard reduction potential data at 25^(@)C is given below E^(@) (Fe^(3+), Fe^(2+)) = +0.77V , E^(@) (Fe^(2+), Fe) = -0.44V , E^(@) (Cu^(2+),Cu) = +0.34V , E^(@)(Cu^(+),Cu) = +0.52 V , E^(@) (O_(2)(g) +4H^(+) +4e^(-) rarr 2H_(2)O] = +1.23V E^(@) [(O_(2)(g) +2H_(2)O +4e^(-) rarr 4OH^(-))] = +0.40V , E^(@) (Cr^(3+), Cr) =- 0.74V , E^(@) (Cr^(2+),Cr) = - 0.91V , Match E^(@) of the redox pair in List-I with the values given in List-II and select the correct answer using the code given below teh lists: {:(List-I,List-II),((P)E^(@)(Fe^(3+),Fe),(1)-0.18V),((Q)E^(@)(4H_(2)O hArr 4H^(+)+4OH^(+)),(2)-0.4V),((R)E^(@)(Cu^(2+)+Curarr2Cu^(+)),(3)-0.04V),((S)E^(@)(Cr^(3+),Cr^(2+)),(4)-0.83V):} Codes:

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Calculate the EMF of the cell Fe(s)+2H^(+)(1M)rarrFe^(+2)(0.001 M)+H_(2)(g)(1 "atm")("given" : E_(Fe^(2+)//Fe)^(@)=-0.44 V)