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The EMF of the cell, Cr|Cr^(+3) (0.1M)...

The `EMF` of the cell,
`Cr|Cr^(+3) (0.1M) ||Fe^(+2) (0.01M)|Fe`
(Given: `E^(0) Cr^(+3)|Cr =- 0.75 V, E^(0) Fe^(+2)|Fe =- 0.45 V)`

Text Solution

AI Generated Solution

To determine the EMF of the cell \(\text{Cr}|\text{Cr}^{3+} (0.1M) || \text{Fe}^{2+} (0.01M)|\text{Fe}\), we will follow these steps: ### Step 1: Write the half-cell reactions **Anode (Oxidation):** \[ \text{Cr} \rightarrow \text{Cr}^{3+} + 3e^- \] ...
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Similar Questions

Explore conceptually related problems

Calculate the emf of the cell Cr|Cr^(3+) (0.1 M) ||Fe^(2+) (0.01 M)|Fe ("Given: "E_(Cr^(3+)//Cr)^(@)=- 0.75" volt, "E_(Fe^(2+)//Fe)^(@)=-0.45" volt")

Calculate the e.m.f. of the cell, Cr//Cr^(3+)(0.1 M) || Fe^(2+)(0.01 M)//Fe "Given" : E_(Cr^(3+)//Cr)^(@)=-0.75" V ", E_(Fe^(2+)//Fe)^(@)=-0.45" V " "Cell reaction" : 2Cr(s)+3Fe^(2+)(aq) to 2Cr^(3+)(aq)+3Fe(s) {"Hint". E_(cell)=E_(cell)^(@)-(0.0591V)/(6)"log"([Cr^(3+)]^(2))/([Fe^(2+)]^(3)}

Knowledge Check

  • 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

    A
    `+0.2606V`
    B
    `+0.5212V`
    C
    `+0.1303V`
    D
    `-0.2606V`.

  • What is the value of E^(@) cell in the following reaction? Cr|Cr^(3+) (0.1 M)||Fe^(2+) (0.01 M)|Fe Given, E_(Cr^(3+)//Cr)^(@)=-0.74 V, E_(Fe^(2+)//Fe)^(@)=-0.44 V

    A
    `+ 0.2606 V`
    B
    `0.5212 V`
    C
    `+ 01303 V`
    D
    `-0.2606 V`

  • What is the e.m.f for the given cell ? Cr | Cr^(3+) (1.0 M) || Co^(2+) (1.0 M) |Co (E^(0) for Cr^(3+) // Cr = -0.74 volt and E^(0) for Co^(2+) // Co = -0.28 volt)

    A
    `-0.46` volt
    B
    `-1.02` volt
    C
    `+0.46` volt
    D
    `1.66` volt

    • Similar Questions

    Explore conceptually related problems

    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 potential of the following half - cells | cells : a. Cr|Cr^(3+)(0.1 M)||Fe^(2+)(0.01M)|Fe Given : E^(c-)._(Cr^(3+)|Cr)=-0.74V E^(c-)._(Fe^(2+)|Fe)=-0.44V b. 6e^(c-)+BrO_(3)^(c-)(aq)+3H_(2)OrarrBr^(c-)(aq)+6OH(aq) Given :E^(c-)._((BrO_(3)^(c-)|Br^(c-)))=0.61V, [BrO_(3)^(c-)]=2.5xx10^(-3)M,[Br^(c-)]=5.0xx10^(-3)M,pH=9.0 c. Ag|Ag^(o+)(0.1M)||Cl^(c-)(0.02M)|Cl_(2)(g)(0.5atm)|Pt Given E_((Ag^(o+)|Ag))=0.80V,E^(c-)((Cl_(2)|2Cl^(c-))=1.36V d. NO_(3) ^(c-)(aq)+2H^(o+)(aq)+e^(-)rarrNO_(2)+H_(2)O Given : E^(c-)._(NO_(3)^(c-)|NO_(2)=0.78V What will be the reductino potential of the half cell in neutral solution ? Assuming all the other species to be at unit concentration.

    Find out work done for the given cell Cr|Cr^(+3) || Fe^(+2)| Fe E_(Cr//Cr^(3+))^(@) = 0.74 V, E_(Fe^(2+)//Fe)^(@) = -0.44 V

    Calculate Delta G^(Theta) and E_(cell) for the cell Al |Al^(3+) (0.01 M) || Fe^(2+) (0.02 M) |Fe Given that E_((Al^(3+)|Al))^(Theta) = -1.66 V and E_((Fe^(2+) |Fe))^(Theta) = -0.44 V

    Calcualte the emf of the following cell at 298 K: 2Cr(s) + 3Fe^(2+) (0.1 M ) to 2Cr^(3+) (0.01 M) + 3Fe(s) Given: E_("Cr^(3+)//Cr)^(@) = -0.74 V, E_(Fe^(2+)//Fe)^(@) = -0.44 V .

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