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Predict reaction of 1N sulphuric acid wi...

Predict reaction of 1N sulphuric acid with following metals : (i) copper (ii) lead (iii) iron Given, `E_(Cu^(2+)|Cu )^(0)`= 0.34volt , `E_(Pb^(2+)|Pb)^(0)` = -0.13 volt, `E_(Fe^(2+)|Fe)^(0)` = -0.44 volt

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To predict the reaction of 1N sulfuric acid with the given metals (copper, lead, and iron), we will analyze the standard reduction potentials (SRP) of each metal compared to hydrogen ions (H⁺) in the acid. The key concept is that metals with a higher SRP than hydrogen will not react with dilute acids to produce hydrogen gas, while those with a lower SRP will. ### Step-by-Step Solution: **Step 1: Understand the standard reduction potentials.** - The standard reduction potentials for the metals are given as follows: - \( E_{Cu^{2+}/Cu}^0 = 0.34 \, \text{V} \) - \( E_{Pb^{2+}/Pb}^0 = -0.13 \, \text{V} \) ...
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Predict reaction of 1 N sulhuric acid with the following metals : (i) copper (ii) lead (iii) iron Given E_(Cu^(2+),Cu)^(@)=+0.34 "volt" ,E_(Pb^(2+),Pb)^(@)=-0.13 "volt" ,E_(Fe^(2+),Fe)^(@)=-0.44 volt

Predict reaction of 1 N sulhuric acid with the following metals : (i) copper (ii) lead (iii) iron Given E_(Cu^(2+),Cu)^(@)=+0.34 "volt" ,E_(Pb^(2+),Pb)^(@)=-0.13 "volt" ,E_(Fe^(2+),Fe)^(@)=-0.44 volt

Knowledge Check

  • E_(Cu^(2+)//Cu)^(@)=0.34V E_(Cu^(+)//Cu)^(@)=0.522V E_(Cu^(2+)//Cu^(+))^(@)=

    A
    0.158
    B
    -0.158
    C
    1.182
    D
    -0.182

  • Copper reduces NO_3^(-) into NO_2 depending upon concentration of HNO_3 in solution Assuming [Cu^(2+)]=0.1M, "and" P_(NO)=P_(NO_2)=10^(-3) bar, at which concentration of HNO_3 , Thermodynamic tendency for reduction of NO_3^(-) into NO and NO_2 by copper is same ? Given: E_(cu^(2+)|cu)^(@)=+0.34 "volt", E_(NO_3^(-)|NO)^(@)=+0.96 "volt",E_(NO_3^(-)|NO_(2))^(@)=+0.76 "volt"]

    A
    `10^(1.23)M`
    B
    `10^(0.56)M`
    C
    `10^(0.66)M`
    D
    `10^(0.12)M

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