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Can a solution of 1 M ZnSO(4) be stored ...

Can a solution of 1 M `ZnSO_(4)` be stored in a vessel made of copper ? Given that
`E_(Zn^(+2)//Zn)^(@) =-0.76V and E_(Cu^(+2)//Cu)^(@)=0.34 V`

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To determine whether a 1 M solution of ZnSO₄ can be stored in a vessel made of copper, we need to analyze the electrochemical properties of zinc and copper based on their standard reduction potentials. ### Step-by-Step Solution: 1. **Identify the Standard Reduction Potentials**: - The standard reduction potential for zinc (Zn²⁺/Zn) is given as \( E^\circ_{Zn^{2+}/Zn} = -0.76 \, V \). - The standard reduction potential for copper (Cu²⁺/Cu) is given as \( E^\circ_{Cu^{2+}/Cu} = +0.34 \, V \). ...
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Calculate the equilibrium constant for the reaction at 298K. Zn(s) +Cu^(2+)(aq) hArr Zn^(2+)(aq) +Cu(s) Given, E_(Zn^(2+)//Zn)^(@) =- 0.76V and E_(Cu^(2+)//Cu)^(@) = +0.34 V

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Calculate the equilibrium constant for the reaction at 298 K Zn(s)+Cu^(2+)(aq)harr Zn^(2+)(aq)+Cu(s) Given " " E_(Zn^(2+)//Zn)^(@)=-0.76 V and E_(Cu^(2+)//Cu)^(@)=+0.34 V

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Calculate the maximum work that can be obtained from the daniell call given below, Zn(s)|Zn^(2+)(aq)||Cu^(2+)(aq)|Cu(s) . Given that E_(Zn^(2+)//Zn)^(@)=-0.76V and E_(Cu^(2+)//Cu)^(@)=+0.34V .

A Galvanic cell consits of three compartment as shown in figure. The first compartment contain ZnSO_4 (1M) and III compartment contain CuSO_4 (1M). The mid compartment contain NaNO_3 (1M). Each compartment contain 1L solution: E_(Zn^(2+)//Zn)^(@)=-0.76 , E_(Cu^(2+)//Cu)^(@)=+0.34 , The concertation of Zn^(2+) in first compartment after passage of 0.1 F charge will be:

A Galvanic cell consits of three compartment as shown in figure. The first compartment contain ZnSO_4 (1M) and III compartment contain CuSO_4 (1M). The mid compartment contain NaNO_3 (1M). Each compartment contain 1L solution: E_(Zn^(2+)//Zn)^(@)=-0.76 , E_(Cu^(2+)//Cu)^(@)=+0.34 The concentration of SO_4^(2-) ions in III compartment after passage of 0.1 F of charge will be:

A Galvanic cell consits of three compartment as shown in figure. The first compartment contain ZnSO_4 (1M) and III compartment contain CuSO_4 (1M). The mid compartment contain NaNO_3 (1M). Each compartment contain 1L solution: E_(Zn^(2+)//Zn)^(@)=-0.76 , E_(Cu^(2+)//Cu)^(@)=+0.34 The concentration of NO_3^- in mid compartment after passage of 0.1 F of charge will be:

What does the negative sign in the expression E_(Zn^(2+)//Zn)^(@)=-0.76V mean?

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