1896 United States presidential election in Maine

1896 United States presidential election in Maine

← 1892 November 3, 1896 1900 →
 
Nominee William McKinley William Jennings Bryan
Party Republican Democratic
Alliance Populist
Home state Ohio Nebraska
Running mate Garret Hobart Arthur Sewall
Electoral vote 6 0
Popular vote 80,403 34,587
Percentage 67.90% 29.21%

County Results
McKinley
  60-70%
  70-80%


President before election

Grover Cleveland
Democratic

Elected President

William McKinley
Republican

The 1896 United States presidential election in Maine took place on November 3, 1896, as part of the 1896 United States presidential election. Voters chose six representatives, or electors to the Electoral College, who voted for president and vice president.

Maine voted for the Republican nominee, former governor of Ohio William McKinley, over the Democratic nominee, former U.S. Representative from Nebraska William Jennings Bryan. McKinley won the state by a margin of 38.69%.

Bryan, running on a platform of free silver, appealed strongly to Western miners and farmers, but had little appeal in the Northeastern states such as Maine.

With 67.90% of the popular vote, Maine would be McKinley's fifth strongest victory in terms of percentage in the popular vote after Vermont, Massachusetts, neighboring New Hampshire and Rhode Island.[1]

Bryan would lose Maine to McKinley again four years later and would later lose the state again in 1908 to William Howard Taft.

  1. ^ "1896 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.