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Determine which element is oxidized and which is reduced CuSO4 plus Mg- Cu plus MgSO4?
In the reaction between CuSO4 and Mg, magnesium (Mg) is oxidized while copper (Cu) is reduced. Magnesium loses electrons to form Mg²⁺ ions, indicating oxidation, while copper ions (Cu²⁺) gain electrons to form elemental copper, indicating reduction. Thus, Mg is the reducing agent and Cu²⁺ is the oxidizing agent in this reaction.
What is the overall voltage for the nonspontaneious redox reaction involving Mg and Cu and their ions?
In the nonspontaneous redox reaction involving magnesium (Mg) and copper ions (Cu²⁺), magnesium acts as the reducing agent, while copper ions are reduced to copper metal. The standard reduction potential for Cu²⁺/Cu is +0.34 V, and for Mg²⁺/Mg, it is -2.37 V. The overall cell potential (E°) can be calculated as E° = E°(reduction) - E°(oxidation), which yields E° = 0.34 V - (-2.37 V) = 2.71 V. Since the reaction is nonspontaneous, the cell potential would be negative under standard conditions.
What is the formula for the ionic compounds formed by magnesium ions and calcium ions with nitrate ions?
The formula for the ionic compound formed by magnesium ions (Mg2+) and nitrate ions (NO3-) is Mg(NO3)2. The formula for the ionic compound formed by calcium ions (Ca2+) and nitrate ions (NO3-) is Ca(NO3)2.
What order of active are H Cu Mg Zn?
H, Mg, Zn, Cu
What is the overall voltage for the nonspontaneous redox reduction involving Mg and Cu and their ions?
The overall voltage for the nonspontaneous redox reaction involving magnesium (Mg) and copper (Cu) can be determined using standard reduction potentials. The reduction potential for Cu²⁺ to Cu is +0.34 V, while the oxidation potential for Mg to Mg²⁺ is -2.37 V. The overall cell potential (E°cell) is calculated by adding the reduction potential of the cathode (Cu) to the oxidation potential of the anode (Mg), resulting in E°cell = 0.34 V - 2.37 V = -2.03 V. Since the value is negative, the reaction is nonspontaneous under standard conditions.
Determine which element is oxidized and which is reduced CuSO4 plus Mg- Cu plus MgSO4?
In the reaction between CuSO4 and Mg, magnesium (Mg) is oxidized while copper (Cu) is reduced. Magnesium loses electrons to form Mg²⁺ ions, indicating oxidation, while copper ions (Cu²⁺) gain electrons to form elemental copper, indicating reduction. Thus, Mg is the reducing agent and Cu²⁺ is the oxidizing agent in this reaction.
What is the overall voltage for the nonspontaneious redox reaction involving Mg and Cu and their ions?
In the nonspontaneous redox reaction involving magnesium (Mg) and copper ions (Cu²⁺), magnesium acts as the reducing agent, while copper ions are reduced to copper metal. The standard reduction potential for Cu²⁺/Cu is +0.34 V, and for Mg²⁺/Mg, it is -2.37 V. The overall cell potential (E°) can be calculated as E° = E°(reduction) - E°(oxidation), which yields E° = 0.34 V - (-2.37 V) = 2.71 V. Since the reaction is nonspontaneous, the cell potential would be negative under standard conditions.
What is the overall voltage for the nonspontaneous redox reaction involving mg and cu and their ions?
-2.37 - 0.34
What is the formula for the ionic compounds formed by magnesium ions and calcium ions with nitrate ions?
The formula for the ionic compound formed by magnesium ions (Mg2+) and nitrate ions (NO3-) is Mg(NO3)2. The formula for the ionic compound formed by calcium ions (Ca2+) and nitrate ions (NO3-) is Ca(NO3)2.
Net inonic equation for Mg plus CuCl2 equals MgCl2 plus Cu?
mg + CuCl2 + MgCl2 + Cu Mg + Cu^+2 = Mg^+2 + Cu
What order of active are H Cu Mg Zn?
H, Mg, Zn, Cu
What is balanced chemical equation of magnesium and copper nitrate?
The balanced chemical equation for the reaction between magnesium and copper(II) nitrate is: Mg + Cu(NO3)2 -> Mg(NO3)2 + Cu
What is the formula of a compound formed when mg plus 2 and cl- react?
MgCl2 is formed. Mg2+ + 2Cl- ----> MgCl2
What is Cu plus MgSO4?
Cu & MgSO4 do NOT react. because in the Reactivity Series magnesium is MORE reactive than copper. So Mg will preferentially oxidise before Cu, Since it is already oxidised in MgSO4 as the ion Mg^(2+), it has already reacted, by ionising two electrons.
Calculate the standard emf of a cell that uses the Mg Mg2 plus and Cu Cu2 plus half cell reactions at 25 degrees C?
Cu + Mg2 --------> Cu2 + Mg Cu --------------> Cu2 + 2e Mg2 + 2e --------> Mg Cu --------------> Cu2 + 2e (E = +0.35) Mg2 + 2e --------> Mg (E = -2.36V) +0.35 + (-2.36) = -2.01V --------------------------------------… Mg + Cu2 --------> Mg2 + Cu Mg --------------> Mg2 + 2e Cu2 + 2e --------> Cu Mg --------------> Mg2 + 2e (E = +2.36V) Cu2 + 2e --------> Mg (E = -0.35V) +2.36 + (-0.35) = +2.01V
Equation for Magnesium plus Copper II nitrate?
Fe + CuSO4 --> FeSO4 + Cu Iron + Copper (II) Sulfate yields Iron (II) Sulfate and Copper
What typically form ions with a plus 2 charge?
The elements that are in group 2 of the Periodic Table, namely Be, Mg, Ca, Sr, Ba and Ra. Some other transition metals will form +2 ions, such as Cu, Zn, Fe, etc.
