1896 United States presidential election in Maryland

1896 United States presidential election in Maryland

← 1892 November 3, 1896 1900 →

Nominee William McKinley William Jennings Bryan Party Republican Democratic Home state Ohio Nebraska Running mate Garret Hobart Arthur Sewall Electoral vote 8 0 Popular vote 136,959 104,150 Percentage 54.73% 41.62%

County Results McKinley   40-50%   50-60%   60-70% Bryan   40-50%   50-60%

President before election Grover Cleveland Democratic Elected President William McKinley Republican

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The 1896 United States presidential election in Maryland took place on November 3, 1896. All contemporary 45 states were part of the 1896 United States presidential election. States voters chose eight electors to the Electoral College, which selected the president and vice president.

Maryland was won by the Republican nominees, former Ohio Governor William McKinley and his running mate Garret Hobart of New Jersey. They defeated the Democratic nominee, former U.S. Representative from Nebraska William Jennings Bryan and his running mate Arthur Sewall. McKinley won the state by a margin of 13.11%.

This was the first time that a Republican presidential candidate won Maryland since Abraham Lincoln in 1864 and the Democrats would not win the state's popular vote until Woodrow Wilson in 1912.

In this election, Maryland voted 8.8% more Republican than the nation at-large.^[1]

Bryan would lose Maryland to McKinley again four years later. In 1908, he would lose the popular vote to William Howard Taft but would win the electoral vote.