United States presidential election in New Jersey, 1860

United States presidential election in New Jersey, 1860

November 6, 1860

Nominee Abraham Lincoln John C. Breckinridge / Stephen A. Douglas Party Republican Fusion Home state Illinois Kentucky / Illinois Running mate Hannibal Hamlin Joseph Lane / Herschel Johnson Electoral vote 4[1] 3 Popular vote 58,346 62,869 Percentage 48.1% 51.9%

President before election James Buchanan Democratic Elected President Abraham Lincoln Republican

Elections in New Jersey
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The 1860 United States presidential election in New Jersey took place on November 6, 1860, as part of the 1860 United States presidential election. Voters chose seven electors of the Electoral College, who voted for president and vice president.

The winner in New Jersey depends on the votes. For the states seven electoral votes, they were split, with Republican candidate Abraham Lincoln carrying four, while the Democratic fusion ticket had Stephen A. Douglas carry three. In the popular vote, the Democratic fusion ticket won the state by a narrow margin of 3.74%.

^{[nb 1]}

New Jersey in the election was one of the four states that had a fusion ticket for the Democratic Party. The other three states were New York, Pennsylvania and Rhode Island.

Results

Notes

References